= Solution
Show that $c$ is a closed <linear subspace> of $\ell^\infty$. Suppose $a^{(j)}\in c$ and $\|a^{(j)}-a\|_\infty\to0$. Given $\varepsilon>0$, choose $j$ with uniform error below $\varepsilon/3$. Since $a^{(j)}$ converges, choose $N$ such that $|a_m^{(j)}-a_n^{(j)}|<\varepsilon/3$ for $m,n\geq N$. Then
$$
|a_m-a_n|\leq|a_m-a_m^{(j)}|+|a_m^{(j)}-a_n^{(j)}|+|a_n^{(j)}-a_n|<\varepsilon.
$$
Thus $(a_n)$ is a real <Cauchy sequence> and converges, so $a\in c$. A closed <linear subspace> of a <Banach space> is complete in the inherited <norm>. Therefore $\boxed{(c,\|\cdot\|_\infty)\text{ is Banach}}$.
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