$\newcommand{\nv}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1\right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}\newcommand{\veps}{\varepsilon}$Let $\{y_n\}_{n=1}^{\infty}$ be a Cauchy sequence in $(X,\nv{\cdot})$, and for every $n\in\mathbf{N}$ write in the Schauder basis, $$y_n=\sum_{i=1}^{\infty}a^{(n)}_ix_i.$$
For any fixed $i\in\mathbf{N}$ that $$|a_i^{(n)}-a_i^{(m)}|\|x_i\|=\|(a_i^{(n)}-a_i^{(m)})x_i\|\leq\left\|\sum_{j=1}^{i}(a_j^{(n)}-a_j^{(m)})x_j\right\|+\left\|\sum_{j=1}^{i-1}(a_j^{(n)}-a_j^{(m)})x_j\right\|\leq 2\nv{y_n-y_m}.$$
Since $x_i\neq 0$, we deduce that $\{a_i^{(n)}\}_{n\in\mathbf{N}}$ is a Cauchy sequence in $\mathbf{R}$, and so $a_i^{(n)}\to a_i$ for some sequence of scalars $a_i$.
Now fix $\veps>0$ . Since $\{y_n\}$ is Cauchy with respect to $\nv{\cdot}$ there exists some $N_0\in\mathbf{N}$ such that for all $n,m\geq N_0$ $\nv{y_n-y_m}<\veps/3$. Now fix any $m\geq N_0$, $M\in\mathbf{N}$, and for all $n\in\mathbf{N}$ define $$z_n^{m,M}=\sum_{i=1}^{M}(a_i^{(n)}-a_i^{(m)})x_i.$$ We see that $\|z_n^{m,M}\|\leq\nv{y_n-y_m}<\veps/3$ for all $n\geq N_0$. Since $z_n^{m,M}\to z^{m,M}:=\sum_{i=1}^{M}(a_i-a_i^{(m)})x_i$ as $n\to\infty$ we see that $\|z^{m,M}\|\leq\veps/3$. Since $M$ was arbitrary we see that for all $m\geq N_0$, \begin{align}
\sup_{M}\left\|\sum_{i=1}^{M}(a_i - a_i^{(m)})x_i\right\|\leq\frac{\veps}{3}.\tag{1}
\end{align}
Now we show that $\sum_{i=1}^{\infty}a_ix_i$ converges in $(X,\|\cdot\|)$. Fix any $m\geq N_0$ and since $y_m\in X$ we see that $\sum_{i=1}^{\infty}a_i^{(m)}x_i$ is convergent in $X$, and so the tail of the sequence can be made arbitrarily small. Rigorously, there exists some $N_1\in\mathbf{N}$ such that $M_2>M_2>N_1$ implies $$\left\|\sum_{i=M_1}^{M_2}a_i^{(m)}x_i\right\|<\frac{\veps}{3}.$$ The triangle inequality gives us $$
\left\|\sum_{i=M_1}^{M_2}a_ix_i\right\| = \left\|\sum_{i=M_1}^{M_2}a_i^{(m)}x_i + \sum_{i=1}^{M_2}(a_i-a_i^{(m)})x_i - \sum_{i=1}^{M_1-1}(a_i-a_i^{(m)})x_i\right\|\leq\veps.$$ Hence $\{\sum_{i=1}^{n}a_ix_i\}_{n\in\mathbf{N}}$ is a Cauchy sequence in $(X,\|\cdot\|)$; since $X$ is complete, $y:=\sum_{i=1}^{\infty}a_ix_i$ converges in $X$.
In light of the above convergence and (1) we see that $\nv{y-y_m}\leq\veps/3$ for all $m\geq N_0$. Since $\veps$ was arbitrary $y_m\to y$ as $m\to\infty$ with respect to $\nv{\cdot}$, which establishes the completeness of $(X,\nv{\cdot})$.
