Here is a complete proof. The following proof is presented in an elementary manner. In fact, no knowledge of topology, or even that $\ell_{\infty}$ can be identified as the dual of $\ell_{1}$, is needed to be able to follow along with the proof in this answer.
Theorem 1. Let $(x_{j})_{j\in\mathbb{N}}$ be a sequence in $\ell_{1}$ and let $x \in \ell_{1}$. Assume that
\begin{equation}
\lim_{j\to\infty} f(x_{j}) = f(x) \tag{1}
\end{equation}
for every $f\in \ell_{1}^{*}$. Then
\begin{equation}
\lim_{j\to\infty} \|x_{j} - x\|_{\ell_{1}} = 0 . \tag{2}
\end{equation}
The proof of Theorem 1 presented in this answer is based on Example 2.5.24 from An Introduction to Banach Space Theory by Megginson. Our plan of attack is to assume for a contradiction that $(2)$ fails. We will then use $(x_{j})_{j\in\mathbb{N}}$ to construct a suitable sequence $(y_{j})_{j\in\mathbb{N}}$ in $\ell_{1}$ such that the condition
\begin{equation}
\lim_{j\to\infty} f(y_{j}) = 0 \quad \text{for all } f \in \ell_{1}^{*}
\end{equation}
both holds and fails to hold, leading to a contradiction.
We also briefly mention the notation used in the statement of Theorem 1 and throughout this answer. The sequence $(x_{j})_{j\in\mathbb{N}}$ is a sequence in $\ell_{1}$. This means that for every $j\in\mathbb{N}$ we have $x_{j} \in \ell_{1}$. We define
\begin{equation}
\ell_{1} = {} \Big\{ y \in \mathbb{K}^{\mathbb{N}} : \sum_{j=1}^{\infty} |y(j)| < \infty \Big\} ,
\end{equation}
where $\mathbb{K}$ is the field over which $\ell_{1}$ is defined over, namely one of $\mathbb{R}$ or $\mathbb{C}$. We define the $\|\cdot\|_{\ell_{1}}$ norm on $\ell_{1}$ by
\begin{equation}
\|y\|_{\ell_{1}} = \sum_{j=1}^{\infty} |y(j)| .
\end{equation}
We now proceed to the proof of Theorem 1. Assume for a contradiction that $(2)$ fails. We will split the proof into three steps.
Step 1. We first show there is a sequence $(y_{j})_{k\in\mathbb{N}}$ in $\ell_{1}$ such that
\begin{equation}
\|y_{j}\|_{\ell_{1}} \geq 1 \tag{3}
\end{equation}
for all $j\in\mathbb{N}$ and that
\begin{equation}
\lim_{j\to\infty} f(y_{j}) = 0 \tag{4}
\end{equation}
for all $f\in \ell_{1}^{*}$. As $(x_{j})_{j\in\mathbb{N}}$ fails to converge to $x$ in the $\ell_{1}$ norm by assumption, we see from a straightforward recursion argument that there is some $\varepsilon > 0$ and a subsequence $(x_{k_{j}})_{j \in \mathbb{N}}$ of $(x_{j})_{j\in\mathbb{N}}$ such that
\begin{equation}
\|x_{k_{j}} - x\|_{\ell_{1}} \geq \varepsilon \tag{5}
\end{equation}
for all $j \in \mathbb{N}$. Define $y_{j} := \varepsilon^{-1} (x_{k_{j}} - x)$ for each $j \in \mathbb{N}$. We have
\begin{equation}
\|y_{j}\|_{\ell_{1}} = \varepsilon^{-1} \|x_{k_{j}} - x\|_{\ell_{1}} \geq \varepsilon^{-1} \cdot \varepsilon = 1
\end{equation}
for each $j\in\mathbb{N}$ by $(5)$ as well as
\begin{equation}
\lim_{j\to\infty} f(y_{j}) = \lim_{j\to\infty} \varepsilon^{-1} (f(x_{k_{j}}) - f(x)) = \varepsilon^{-1} (f(x) - f(x)) = 0
\end{equation}
for each $f\in \ell_{1}^{*}$ by $(1)$ and the linearity of $f$. This completes Step 1.
Step 2. We show given the sequence $(y_{j})_{j\in\mathbb{N}}$ from Step 1, that by passing to a subsequence if necessary, we may further assume there is a strictly increasing sequence $(n_{j})_{j\in\mathbb{N}}$ in $\mathbb{N}_{0}$ with $n_{1} = 0$ such that
\begin{equation}
\sum_{m = n_{j} + 1}^{n_{j+1} } |y_{j}(m)| > \frac{3}{4} \|y_{j} \|_{\ell_{1}} \tag{6}
\end{equation}
for all $j\in\mathbb{N}$. We first argue by recursion that we can find a strictly increasing sequence $(n_{j})_{j\in\mathbb{N}}$ in $\mathbb{N}_{0}$ with $n_{1} = 0$ and a strictly increasing sequence $(k_{j})_{j\in\mathbb{N}}$ in $\mathbb{N}$ with $k_{1} = 1$ such that
\begin{equation}
\sum_{m = n_{j} + 1}^{n_{j+1} } |y_{k_{j}}(m)| > \frac{3}{4} \|y_{k_{j}} \|_{\ell_{1}} \quad \text{and} \quad \sum_{m = 1}^{n_{j} } |y_{k_{j}}(m)| < \frac{1}{4} \tag{7}
\end{equation}
for all $j\in\mathbb{N}$. We use the convention that $\sum_{m=j}^{0} a_{m} = 0$ for any $j\in\mathbb{N}$ and any sequence $(a_{m})_{m\in\mathbb{N}}$ in $\mathbb{C}$.
We now define the sequence recursively. We set $n_{1} = 0$ and $k_{1} = 1$. Let $j\in\mathbb{N}$ and assume that strictly increasing finite sequences $(n_{i})_{i=1}^{j}$ and $(k_{i})_{i=1}^{j}$ in $\mathbb{N}_{0}$ and $\mathbb{N}$ respectively are given such that
\begin{equation}
\sum_{m = n_{i} + 1}^{n_{i+1} } |y_{k_{i}}(m)| > \frac{3}{4} \|y_{k_{i}} \|_{\ell_{1}}
\end{equation}
for all $i\in \mathbb{N}$ with $i < j$ and
\begin{equation}
\sum_{m = 1}^{n_{i} } |y_{k_{i}}(m)| < \frac{1}{4} \tag{8}
\end{equation}
for all $i\in \mathbb{N}$ with $i \leq j$. Applying $(8)$ with $i = j$ as well as using $\|y_{k_{j}} \|_{\ell_{1}} \geq 1$ from $(3)$, we see that
\begin{equation}
\sum_{m = n_{j} + 1}^{\infty} |y_{k_{j}}(m)| = \|y_{k_{j}} \|_{\ell_{1}} - \sum_{m = 1}^{n_{j} } |y_{k_{j}}(m)| > \|y_{k_{j}} \|_{\ell_{1}} - \frac{1}{4} \geq \|y_{k_{j}} \|_{\ell_{1}} - \frac{1}{4} \|y_{k_{j}} \|_{\ell_{1}} = \frac{3}{4} \|y_{k_{j}} \|_{\ell_{1}} .
\end{equation}
Hence $\sum_{m = n_{j} + 1}^{\infty} |y_{k_{j}}(m)| > \tfrac{3}{4} \|y_{k_{j}} \|_{\ell_{1}}$. Because
\begin{equation}
\sum_{m = n_{j} + 1}^{\infty} |y_{k_{j}}(m)| = \sup\Big\{\sum_{m = n_{j} + 1}^{N} |y_{k_{j}}(m)| : N\in\mathbb{N} \text{ and } N > n_{j} \Big\} ,
\end{equation}
there is some $n_{j+1} \in \mathbb{N}$ with $n_{j} < n_{j+1}$ such that
\begin{equation}
\sum_{m = n_{j} + 1}^{n_{j+1} } |y_{k_{j}}(m)| > \frac{3}{4} \|y_{k_{j}} \|_{\ell_{1}} .
\end{equation}
Let $m\in \mathbb{N}$ with $n_{1} < m \leq n_{j+1}$. As the map $y\mapsto y(m)$ defined on $\ell_{1}$ is a member of $\ell_{1}^{*}$, it follows from $(4)$ that
\begin{equation}
\lim_{j\to\infty} y_{j}(m) = 0
\end{equation}
and hence
\begin{equation}
\lim_{j\to\infty} |y_{j}(m)| = 0 . \tag{9}
\end{equation}
As $m$ was arbitrary, we deduce that $(9)$ holds for all $m\in\mathbb{N}$ with $n_{1} < m \leq n_{j+1}$. This implies that
\begin{equation}
\lim_{j\to\infty} \sum_{m = n_{1} + 1}^{n_{j+1} } |y_{j}(m)| = 0 .
\end{equation}
In particular, there is some $k_{j+1} \in \mathbb{N}$ with $k_{j} < k_{j+1}$ such that
\begin{equation}
\sum_{m = n_{1} + 1}^{n_{j+1} } |y_{k_{j+1}}(m)| < \frac{1}{4} .
\end{equation}
Hence we have chosen $n_{j+1}$ and $k_{j+1}$ with the desired properties. We now conclude by recursion that we can find a strictly increasing sequence $(n_{j})_{j\in\mathbb{N}}$ in $\mathbb{N}_{0}$ with $n_{1} = 0$ and a strictly increasing sequence $(k_{j})_{j\in\mathbb{N}}$ in $\mathbb{N}$ with $k_{1} = 1$ such that $(7)$ holds. The sequence $(y_{k_{j}})_{j\in\mathbb{N}}$ combined with the strictly increasing sequence $(n_{j})_{j\in\mathbb{N}}$ have the desired properties. This completes Step 2.
Step 3. We show there is some $f\in \ell_{1}^{*}$ such that
\begin{equation}
|f(y_{j})| \geq \frac{1}{2} \tag{10}
\end{equation}
for all $j\in\mathbb{N}$. Note that once we have shown this, we obtain a contradiction when combining $(10)$ with $(4)$ and our result follows.
Let $(n_{j})_{j\in\mathbb{N}}$ be a strictly increasing sequence in $\mathbb{N}_{0}$ with $n_{1} = 0$ such that $(6)$ holds for all $j\in\mathbb{N}$. For each $m\in\mathbb{N}$ let $\gamma_{m}\in\mathbb{K}$ be a scalar with $|\gamma_{m}| = 1$ such that if $j$ is the unique member of $\mathbb{N}$ for which $n_{j} < m \leq n_{j+1}$, then $\gamma_{m} y_{j}(m) = |y_{j}(m)|$. Define $f\colon \ell_{1} \to \mathbb{K}$ by
\begin{equation}
f(y) := \sum_{m=1}^{\infty} \gamma_{m} y(m) .
\end{equation}
Since $(\gamma_{m})_{m\in\mathbb{N}}$ is a bounded sequence, it is straightforward to check that $f$ is well-defined and $f\in \ell_{1}^{*}$.
Fix $j\in\mathbb{N}$. Note that by applying $(6)$ twice we have
\begin{equation}
\begin{aligned}
\Big| \sum_{m = 1}^{n_{j}} \gamma_{m} y_{j}(m) + \sum_{m = n_{j+1} + 1}^{\infty} \gamma_{m} y_{j}(m) \Big| &\leq {} \Big| \sum_{m = 1}^{n_{j}} \gamma_{m} y_{j}(m) \Big| + {} \Big| \sum_{m = n_{j+1} + 1}^{\infty} \gamma_{m} y_{j}(m) \Big| \\
&\leq \sum_{m = 1}^{n_{j}} |y_{j}(m)| + \sum_{m = n_{j+1} + 1}^{\infty} |y_{j}(m)| \\
&= \|y_{j}\|_{\ell_{1}} - \sum_{m = n_{j} + 1}^{n_{j+1}} |y_{j}(m)| \\
&< \|y_{j}\|_{\ell_{1}} - \frac{3}{4} \|y_{j}\|_{\ell_{1}} \\
&= \frac{1}{4} \|y_{j}\|_{\ell_{1}}
\end{aligned} \tag{11}
\end{equation}
and
\begin{equation}
\Big| \sum_{m = n_{j} + 1}^{n_{j+1}} \gamma_{m} y_{j}(m) \Big| \geq {} {\rm Re} \Big( \sum_{m = n_{j} + 1}^{n_{j+1}} \gamma_{m} y_{j}(m) \Big) = \sum_{m = n_{j} + 1}^{n_{j+1}} {\rm Re} (\gamma_{m} y_{j}(m)) = \sum_{m = n_{j} + 1}^{n_{j+1}} |y_{j}(m)| > \frac{3}{4} \|y_{j} \|_{\ell_{1}} . \tag{12}
\end{equation}
By combining $(11)$ and $(12)$ as well as recalling that $\|y_{j}\|_{\ell_{1}} \geq 1$ and applying the reverse triangle inequality, it follows that
\begin{align*}
|f(y_{j})| &= {} \Big| \sum_{m=1}^{\infty} \gamma_{m} y_{j}(m) \Big| \\
&= {} \Big| \sum_{m=1}^{n_{j}} \gamma_{m} y_{j}(m) + \sum_{m = n_{j} + 1}^{n_{j+1}} \gamma_{m} y_{j}(m) + \sum_{m = n_{j+1} + 1}^{\infty} \gamma_{m} y_{j}(m) \Big| \\
&\geq {} \Big| \sum_{m = n_{j} + 1}^{n_{j+1}} \gamma_{m} y_{j}(m) \Big| - {} \Big| \sum_{m = 1}^{n_{j}} \gamma_{m} y_{j}(m) + \sum_{m = n_{j+1} + 1}^{\infty} \gamma_{m} y_{j}(m) \Big| \\
&\geq \frac{3}{4} \|y_{j} \|_{\ell_{1}} - \frac{1}{4} \|y_{j} \|_{\ell_{1}} \\
&= \frac{1}{2} \|y_{j} \|_{\ell_{1}} \\
&\geq \frac{1}{2} .
\end{align*}
As $j\in\mathbb{N}$ was arbitrary, this shows that $(10)$ holds for all $j\in\mathbb{N}$. But $(10)$ contradicts $(4)$. As our assumption that $(1)$ fails lead us to a contradiction, we conclude that $(1)$ does hold as claimed in the statement of Theorem 1. This completes Step 3 and finishes the proof.
Further information
Here are a few other ways of proving Theorem 1 with references to proofs included.
See Theorem 5.36 of Banach Space Theory: The Basis for Linear and Nonlinear Analysis by Fabian et al. for a proof similar to the one in this answer. The proof in Fabian et al. also assumes for a contradiction that there is a sequence that converges to $0$ in the weak topology but not in the norm topology and then deduces a contradiction, but the presentation has some differences to the one in this answer.
See Proposition V.5.2 of A Course in Functional Analysis by Conway for a proof that makes use of the Baire category theorem.
See the corollary in Section V.1 of Functional Analysis by Yosida for a proof that makes use of the Vitali-Hahn-Saks theorem. Also see this answer.
See Schur's Theorem in Chapter VII of Sequences and Series in Banach Spaces by Diestel for a proof that makes use of Phillips's lemma.
Finally, note that the proof in this answer can be modified without too much work to prove a more general statement. We first need to cover some background. Let $\Gamma$ be a set and $\mathbb{K}$ a field fixed as one of $\mathbb{R}$ or $\mathbb{C}$. For each $\gamma \in \Gamma$ let $X_{\gamma}$ be a Banach space over $\mathbb{K}$. We define
\begin{equation}
(\oplus_{\gamma \in \Gamma} X_{\gamma} )_{1} = {} \Big\{ (x_{\gamma})_{\gamma \in \Gamma} : x_{\gamma} \in X_{\gamma} \text{ for each } \gamma \in \Gamma \text{ and } \sup\Big\{\sum_{\gamma \in F} \|x_{\gamma}\|_{X_{\gamma}} : F\subseteq \Gamma \text{ finite} \Big\} < \infty \Big\}
\end{equation}
equipped with the complete norm
\begin{equation}
\|(x_{\gamma})_{\gamma \in \Gamma} \|_{1} = \sup\Big\{\sum_{\gamma \in F} \|x_{\gamma} \|_{X_{\gamma}} : F \subseteq \Gamma \text{ finite} \Big\} .
\end{equation}
It is routine to check that $(\oplus_{\gamma \in \Gamma} X_{\gamma} )_{1}$ is indeed a vector space over $\mathbb{K}$ and that $\| \cdot \|_{1}$ is indeed a complete norm on $(\oplus_{\gamma \in \Gamma} X_{\gamma} )_{1}$. Hence $(\oplus_{\gamma \in \Gamma} X_{\gamma} )_{1}$ is a Banach space when equipped with the norm $\| \cdot \|_{1}$.
Let $X$ be a Banach space and $x\in X$. We say a sequence $(x_{j})_{j\in\mathbb{N}}$ in $X$ is weak convergent to $x$ if
\begin{equation}
\lim_{j\to\infty} f(x_{j}) = f(x)
\end{equation}
for all $f\in X^{*}$, while we say $(x_{j})_{j\in\mathbb{N}}$ is norm convergent to $x$ if
\begin{equation}
\lim_{j\to\infty} \|x_{j} - x\|_{X} = 0 .
\end{equation}
It is straightforward to see that every norm convergent sequence in $X$ is weak convergent. The converse typically fails. We say that $X$ has Schur's property if every weak convergent sequence in $X$ is norm convergent in $X$. Note that Theorem 1 says that $\ell_{1}$ has Schur's property. We have the following more general result.
Theorem 2. Let $\Gamma$ be a set and $\mathbb{K}$ a field fixed as one of $\mathbb{R}$ or $\mathbb{C}$. For each $\gamma \in \Gamma$ let $X_{\gamma}$ be a Banach space over $\mathbb{K}$ with Schur's property. Then $(\oplus_{\gamma \in \Gamma} X_{\gamma} )_{1}$ is a Banach space with Schur's property.
Sketch of Proof. We first reduce the problem to the case where $\Gamma = \mathbb{N}$. Let $(x_{j})_{j\in\mathbb{N}}$ be a sequence in $(\oplus_{\gamma \in \Gamma} X_{\gamma} )_{1}$ which is weak convergent to $0$. For each $j\in\mathbb{N}$ we have that $\{\gamma \in \Gamma : x_{j}(\gamma ) \neq 0 \}$ is countable. This is readily seen from the definition of $(\oplus_{\gamma \in \Gamma} X_{\gamma} )_{1}$. Hence the set
\begin{equation}
\Gamma_{0} := \{\gamma \in \Gamma : x_{j}(\gamma ) \neq 0 \text{ for some } j \in\mathbb{N} \}
\end{equation}
is countable. Now it is readily seen that $(\oplus_{\gamma \in \Gamma_{0}} X_{\gamma} )_{1}$ is isometrically isomorphic to a closed subspace of $(\oplus_{\gamma \in \Gamma} X_{\gamma} )_{1}$ containing the sequence $(x_{j})_{j\in\mathbb{N}}$. Using this isomorphism we can restrict our attention to the case where $\Gamma$ is countable. Furthermore, if $\Gamma$ is finite then the result follows from comparing the finite number of components. Hence we may assume $\Gamma$ is countably infinite and, by using a bijection if necessary, that $\Gamma = \mathbb{N}$.
The rest of the proof proceeds similarly to the proof of Theorem 1 with proof by contradiction along the same three steps. The first two steps carry through to this setting with essentially no nontrivial modifications. For the third step, the only small change that needs to be made is to replace the use of the scalars $(\gamma_{m})_{m\in\mathbb{N}}$ from the Step 3 of Theorem 1 with well-chosen bounded linear functionals. In more detail, if $(y_{j})_{j\in\mathbb{N}}$ is a sequence in $(\oplus_{m\in\mathbb{N}} X_{m} )_{1}$ that is weak convergent to $0$ but satisfies $\|y_{j}\| \geq 1$ for each $j\in\mathbb{N}$, and there is also a strictly increasing sequence $(n_{j})_{j\in\mathbb{N}}$ in $\mathbb{N}_{0}$ such that
\begin{equation}
\sum_{m = n_{j} + 1}^{n_{j+1} } \|y_{j}(m)\|_{X_{m}} > \frac{3}{4} \|y_{j} \|_{1}
\end{equation}
for all $j\in\mathbb{N}$, then for each $m\in\mathbb{N}$ one takes $x_{m}^{*} \in X_{m}^{*}$ with $\| x_{m}^{*} \|_{X_{m}^{*}} = 1$ such that if $j$ is the unique member of $\mathbb{N}$ for which $n_{j} < m \leq n_{j+1}$, then $x_{m}^{*}(y_{j}(m)) = \|y_{j}(m) \|_{X_{m}}$. We then define $f\colon (\oplus_{m\in\mathbb{N}} X_{m} )_{1} \to \mathbb{K}$ by
\begin{equation}
f(y) := \sum_{m=1}^{\infty} x_{m}^{*}(y(m) ) .
\end{equation}
One observes that $f$ is well-defined and $f\in (\oplus_{m\in\mathbb{N}} X_{m} )_{1}^{*}$. Finally, one proceeds as in the remaining part of Step 3 of Theorem 1 to deduce that
\begin{equation}
|f(y_{j})| \geq \frac{1}{2}
\end{equation}
for all $j\in\mathbb{N}$ and deduce the desired contradiction, which proves Theorem 2.
