I proved in class the space of bounded sequences with the same metric is complete, so I shall use this fact. How's my proof? BTW, regarding notation, let $x_{jk}$ refer to the $k$th element in the $j$th sequence.
Consider an arbitrary Cauchy sequence of bounded sequences converging to zero $\{x_1\}, \{x_2\}, \cdots$. We know by part (a) that such a sequence has a limit, let's call it $\{x\}$. We just need to show that $\{x\}$ converges to 0. We know by definition of convergence that for an error of $1/2m$, there exists an $N$ such that for $k \geq N, d(\{x_k\},\{x\}) = sup_i|x_{ki}-x_i| \leq 1/2m \Longleftrightarrow \forall i, |x_{ki}-x_i| \leq 1/2m$. Fixing a $k \geq N$, we know that for $\{x_k\}$, because this sequence converges to 0, there exists a $J$ such that for all $j \geq J$, $|x_{kj}| < 1/2m$. Thus, for $j \geq J$, we have $$|x_j| \leq |x_j-x_{kj}| + |x_{kj}| \leq 1/2m + 1/2m = 1/m$$
Ok so you want to show that every Cauchy sequence $\{[x_n]\}$ in the subspace of bounded sequences with limit zero (let's call it $c_0$) has a limit in that space. Now, what does it mean to be in $c_0$? It means that given $[x] \in c_0$, for every $\epsilon>0$ there exists $N>0$ such that $|x ^i| \leq \epsilon$ for every $i \geq N$. You write that you already proved completeness of $\ell ^\infty$, so consider $[x]$ the limit of $\{ [x_n] \}$ in $\ell ^\infty$. If $[x]$ didn't belong to $c_0$ then $\exists \epsilon_0 > 0$ such that $|x^{i(k)} |> \epsilon _0$ for an increasing sequence of $i(k)$. Now, since $[x_n] \to [x]$ in $ \ell^\infty$, $\exists $N such that $d([x],[x_n]) \leq \epsilon_0 /2$ for every $n \geq N$. But then we have an absurd. Indeed if the previous were true, then for every $k$:
$$ |x_n ^{i(k)}| \geq | (|x^{i(k)}|- |x_n ^{i(k)} - x^{i(k)}| )|\geq \epsilon_0 - \epsilon_0 /2= \epsilon_0 /2$$
by the reverse triangle inequality, which is absurd since $[x_n] \in c_0$.