= Solution
Yes. Let $(x^{(k)})_{k\geq1}$ be any <sequence> in $A$, where $x^{(k)}=(x_m^{(k)})_{m\geq1}$. At coordinate $m$, the values lie in the <compact interval> $[-1/m,1/m]$. By part (a)(ii) we can successively extract a <convergent subsequence> for coordinate one, a further one for coordinate two, and so on. Taking the diagonal indices gives a single subsequence $(x^{(k_j)})$ for which
$$
x_m^{(k_j)}\longrightarrow x_m\quad\text{for every fixed }m.
$$
The <closed interval> bound passes to the limit, so $|x_m|\leq1/m$. Hence $x_m\to0$ and $x=(x_m)\in A$.
Coordinate convergence alone would not imply <supremum norm> convergence. Here the common tail bound supplies the missing estimate. Given $\epsilon>0$, choose $M$ with $2/(M+1)<\epsilon$. For $m>M$,
$$
|x_m^{(k_j)}-x_m|\leq2/m\leq2/(M+1)<\epsilon.
$$
For the finitely many $m\leq M$, coordinate convergence makes all differences less than $\epsilon$ once $j$ is large. Therefore $d(x^{(k_j)},x)<\epsilon$, and
$$
\boxed{A\text{ is sequentially compact in the supremum norm metric}}.
$$
This is an instance of <compactness from uniformly vanishing sequence tails>. The common $1/m$ bound is exactly what the coordinate vectors in part (i) lack.
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