Note that part (a) (where the typo from Albiac and Kalton's book involving the strict inequality is replaced with an inequality) is just one formulation of a result named Helly's theorem. See here for a proof.
Part (b) takes some work. We will make use of the following lemma.
Lemma 1. Let $(y_{n})_{n\in\mathbb{N}}$ be a block basic sequence taken from the standard basis of $\ell_{1}$. Then $(\|y_{n}\|^{-1} y_{n} )_{n\in\mathbb{N}}$ is a basic sequence equivalent to the standard basis of $\ell_{1}$ and there is a bounded linear projection on $\ell_{1}$ with range $\overline{{\rm span}}\{y_{n} : n\in\mathbb{N} \}$.
Proof. See Lemma 2.1.1 from Topics in Banach Space Theory by Albiac and Kalton.
We begin the proof regarding the question about part (b). We assume that $X^{*}$ has a closed subspace isomorphic to $c_{0}$ and show that $X$ has a complemented closed subspace isomorphic to $\ell_{1}$. By assumption we have a bounded linear operator $T\colon c_{0} \to X^{*}$ which is an isomorphism onto its image. For each $n\in\mathbb{N}$ define $e_{n} \colon \mathbb{N} \to \mathbb{K}$ by $e_{n}(k) = 1$ if $k = n$ and $0$ otherwise. We have that $(e_{n})_{n\in\mathbb{N}}$ is the standard basis for both $c_{0}$ and $\ell_{1}$. Let $(e^{*}_{n})_{n\in\mathbb{N}}$ be the basis for $c_{0}^{*}$ such that $e^{*}_{n}(e_{m}) = 1$ if $m = n$ and $0$ otherwise for $m,n\in\mathbb{N}$. Let $J$ be the canonical embedding of $X$ into $X^{**}$. For each $n\in\mathbb{N}$ define
\begin{equation}
V_{n} := {} \Big\{ y^{*} \in c_{0}^{*} : |y^{*}(e_{m})| < \frac{1}{n} \text{ for all } m\in \mathbb{N} \text{ with } 1 \leq m < n , |y^{*}(e_{n})| > \frac{1}{2} \Big\} . \tag{1}
\end{equation}
We wish to find a bounded sequence $(x_{n})_{n\in\mathbb{N}}$ in $X$ such that $T^{*}(Jx_{n}) \in V_{n}$ for all $n\in\mathbb{N}$ (done in Step 1), then apply the Bessaga-Pełcyński selection principle to this sequence to obtain a suitable subsequence which incudes a basic sequence equivalent to the standard basis for $\ell_{1}$ (done in Step 2) and then construct our desired projection from there (Steps 3 and 4).
Step 1. We first show a sequence $(x_{n})_{n\in\mathbb{N}}$ exists which satisfies the following conditions.
$({\rm i})$ The sequence $(x_{n})_{n\in\mathbb{N}}$ is bounded in $X$.
$({\rm ii})$ We have $|(T^{*}Jx_{n})(e_{m})| < \tfrac{1}{n}$ for all $m, n \in\mathbb{N}$ with $1 \leq m < n$.
$({\rm iii})$ We have $|(T^{*}Jx_{n})(e_{n})| > \tfrac{1}{2}$ for all $n\in\mathbb{N}$.
As $T$ is an isomorphism onto its image, we have that $T^{*}\colon X^{**} \to c_{0}^{*}$ is a surjective bounded linear operator (the range of $T^{*}$ is weak$^{*}$ dense as $T$ is injective and is further weak$^{*}$ closed by the closed range theorem). By the open mapping theorem we have that $T^{*}(B_{X^{**}})$ is a neighbourhood of $0$ in $c_{0}^{*}$. Hence there is some $\lambda > 0$ such that $B_{c_{0}^{*}} \subseteq T^{*}(\lambda B_{X^{**}})$. By Goldstine's theorem we have that $J(\lambda B_{X})$ is weak$^{*}$ dense in $\lambda B_{X^{**}}$. It follows from the weak$^{*}$-weak$^{*}$ continuity of $T^{*}$ as an adjoint operator that
\begin{equation}
B_{c_{0}^{*}} \subseteq T^{*}(\lambda B_{X^{**}}) \subseteq \overline{T^{*}(J(\lambda B_{X}))}^{{\rm weak}^{*}} . \tag{2}
\end{equation}
Now let $n\in\mathbb{N}$. We have that $V_{n}$ is a neighbourhood of $e^{*}_{n}$ in $(c_{0}^{*}, {\rm weak}^{*})$, so by $(2)$ there is some $x_{n} \in \lambda B_{X}$ such that $T^{*}(Jx_{n})) \in V_{n}$. As $n\in\mathbb{N}$ is arbitrary and $\lambda > 0$ is independent of the choice of $n$, the construction of the sequence $(x_{n})_{n\in\mathbb{N}}$ follows.
Step 2. We use the Bessaga-Pełcyński selection principle to find a subsequence $(x_{k_{n}})_{n\in\mathbb{N}}$ of $(x_{n})_{n\in\mathbb{N}}$ and then use this subsequence to obtain a basic sequence equivalent to the standard basis of $\ell_{1}$. Let $\Lambda \colon c_{0}^{*} \to \ell_{1}$ be the standard isomorphism between $c_{0}^{*}$ and $\ell_{1}$ given by $(\Lambda y^{*})(n) := y^{*}(e_{n})$. Define $S := \Lambda T^{*} J$. Then $S$ is a bounded linear operator that maps $S$ into $\ell_{1}$. Observe that for all $m,n\in\mathbb{N}$ we have
\begin{equation}
(Sx_{n})(m) = (\Lambda (T^{*}Jx_{n}) )(m) = (T^{*}Jx_{n})(e_{m}) .
\end{equation}
It now follows from $({\rm ii})$ of Step 1 that
\begin{equation}
\lim_{n\to\infty} |(Sx_{n})(m)| = 0
\end{equation}
for all $m\in\mathbb{N}$ and from $({\rm iii})$ of Step 1 that
\begin{equation}
\| Sx_{n} \| > \tfrac{1}{2}
\end{equation}
for all $n\in\mathbb{N}$. By the Bessaga-Pełcyński selection principle we deduce there is a subsequence $(x_{k_{n}})_{n\in\mathbb{N}}$ of $(x_{n})_{n\in\mathbb{N}}$ such that $(Sx_{k_{n}})_{n\in\mathbb{N}}$ is a basic sequence in $\ell_{1}$ equivalent to a block basic sequence $(z_{n})_{n\in\mathbb{N}}$ in $\ell_{1}$ taken from $(e_{n})_{n\in\mathbb{N}}$.
By Lemma 1 we have that $(\|z_{n}\|^{-1} z_{n} )_{n\in\mathbb{N}}$ is a complemented basic sequence in $\ell_{1}$ equivalent to the standard basis of $\ell_{1}$. It follows that $(\|z_{n}\|^{-1} Sx_{k_{n}})_{n\in\mathbb{N}}$ is a complemented basic sequence in $\ell_{1}$ equivalent to $(\|z_{n}\|^{-1} z_{n} )_{n\in\mathbb{N}}$ and hence also to the standard basis of $\ell_{1}$. As $(e_{n})_{n\in\mathbb{N}}$ is bounded in $\ell_{1}$, it follows that $(\|z_{n}\|^{-1} Sx_{k_{n}} )_{n\in\mathbb{N}}$ is also bounded and so there is some $c_{1} > 0$ such that
\begin{equation}
c_{1} \geq {} \big\| \|z_{n}\|^{-1} Sx_{k_{n}} \big\| = \| Sx_{k_{n}} \| \|z_{n}\|^{-1} > \tfrac{1}{2} \|z_{n}\|^{-1}
\end{equation}
for all $n\in\mathbb{N}$. In particular, we have $1 \leq 2c_{1} \|z_{n}\|$ for all $n\in\mathbb{N}$.
Step 3. Define $Y := \overline{{\rm span}}\{x_{k_{n}} : n\in\mathbb{N} \}$ and $Z := \overline{{\rm span}}\{\|z_{n}\|^{-1} Sx_{k_{n}} : n\in\mathbb{N} \}$. We have that $Z$ is isomorphic to $\ell_{1}$ because $(\|z_{n}\|^{-1} Sx_{k_{n}})_{n\in\mathbb{N}}$ is a basic sequence in $\ell_{1}$ equivalent to the standard basis of $\ell_{1}$. We claim that $S$ is an isomorphism from $Y$ onto $Z$. Recalling the sequence $(x_{k_{n}})_{n\in\mathbb{N}}$ is bounded by $({\rm i})$ from Step 1, there is some $c_{2} > 0$ such that
\begin{equation}
\|x_{k_{n}} \| \leq c_{2}
\end{equation}
for all $n\in\mathbb{N}$. Furthermore, because the basic sequence $(\|z_{n}\|^{-1} Sx_{k_{n}} )_{n\in\mathbb{N}}$ is equivalent to the standard basis of $\ell_{1}$, there is some $c_{3} > 0$ such that for any $m\in\mathbb{N}$ and $a_{1}, \ldots , a_{m} \in \mathbb{K}$ we have
\begin{equation}
\Big\| \sum_{n=1}^{m} a_{n} e_{n} \Big\| \leq c_{3} \Big\| \sum_{n=1}^{m} a_{n} \|z_{n}\|^{-1} Sx_{k_{n}} \Big\| .
\end{equation}
Let $m\in\mathbb{N}$ and $a_{1}, \ldots , a_{m} \in \mathbb{K}$. Using that $1 \leq 2c \|z^{*}_{n}\|$ for each $n\in\mathbb{N}$, we have
\begin{equation}
\begin{aligned}
\Big\| \sum_{n=1}^{m} a_{n} x_{k_{n}} \Big\| &\leq c_{2} \sum_{n=1}^{m} |a_{n}| \\
&\leq c_{2} \sum_{n=1}^{m} 2 c_{3} \|z_{n}\| |a_{n}| \\
&= 2 c_{2} c_{3} \Big\| \sum_{n=1}^{m} \|z_{n}\| a_{n} e_{n} \Big\| \\
&= 2 c_{2} c_{3} \Big\| \sum_{n=1}^{m} \|z_{n}\| a_{n} (\|z_{n}\|^{-1} Sx_{k_{n}}) \Big\| \\
&= 2 c_{2} c_{3} \Big\| S \big( \sum_{n=1}^{m} a_{n} x_{k_{n}} \big) \Big\| .
\end{aligned} \tag{3}
\end{equation}
It follows from $(3)$ combined with the continuity of $S$ that $S$ is an isomorphism from $Y$ onto $Z$. One consequence is that as $Z$ is isomorphic to $\ell_{1}$, this means that $Y$ is also isomorphic to $\ell_{1}$. Another consequence is that there is a bounded linear operator $R\colon Z\to X$ such that ${\rm ran} \, R = Y$ and $SRy = y$ for all $y\in Z$.
Step 4. We now construct a bounded linear projection on $X$ with range $Y$. Since $(\|z_{n}\|^{-1} Sx_{k_{n}})_{n\in\mathbb{N}}$ is a complemented basic sequence in $\ell_{1}$ there is a bounded linear operator $P\colon \ell_{1} \to Z$ such that $Py = y$ for all $y\in Z$. Define $Q := RPS$. Note that ${\rm ran} \, PS = Z$ and hence ${\rm ran} \, RPS = {\rm ran} \, R = Y$. Since $PSR = {\rm id}_{Y}$ we have
\begin{equation}
Q^{2} = (RPS)^{2} = R (PSR) PS = RPS = Q .
\end{equation}
Hence $Q$ is a bounded linear projection with range $Y$. Therefore $X$ has a complemented closed subspace $Y$ isomorphic to $\ell_{1}$.
Step 5. We deduce that $X^{*}$ contains a closed subspace isomorphic to $\ell_{\infty}$. Although the author of the question is already aware of how to show this, it is included for completeness for future readers. Recall from Step 4 that there is a bounded linear projection $Q$ on $X$ whose range is isomorphic to $\ell_{1}$. It is straightforward to check that the adjoint operator $Q^{*}$ is a weak$^{*}$-weak$^{*}$ continuous projection on $X^{*}$. Hence ${\rm ran} \, Q^{*}$ is weak$^{*}$ closed in $X^{*}$ and we have ${\rm ran} \, Q^{*} = ({\rm ker} \, Q)^{\perp}$. It remains to show that ${\rm ran} \, Q^{*}$ is isomorphic to $({\rm ran} \, Q)^{*}$. Let $\Psi \colon X\to {\rm ran} \, Q$ be defined by $\Psi x := Qx$. We claim that $\Psi^{*}$ is an isomorphism from $({\rm ran} \, Q)^{*}$ onto ${\rm ran} \, Q^{*}$. As it is straightforward to check that $\Psi$ is injective and is a weak$^{*}$-weak$^{*}$ homeomorphism onto its image, it suffices to check that ${\rm ran} \, \Phi^{*} = {\rm ran} \, Q^{*}$. This follows from noting that
\begin{equation}
{\rm ran} \, \Phi^{*} = \{g\circ \Phi : g\in ({\rm ran} \, \Phi )^{*} \} = \{f\circ Q : f\in ({\rm ran} \, Q )^{*} \} = {\rm ran} \, Q^{*} .
\end{equation}
Now recalling that ${\rm ran} \, Q$ is isomorphic to $\ell_{1}$ and $\ell_{1}^{*}$ is isomorphic to $\ell_{\infty}$, we have that $Q^{*}$ is a weak$^{*}$-weak$^{*}$ continuous projection $X^{*}$ whose range is isomorphic to a closed subspace of $\ell_{\infty}$. This completes the proof.
