(Converse of my last question)
If $A \subseteq \ell_\infty$, and $A=\{l\in \ell_\infty: |l_n| \le b_n \}$, where $b_n$ is a sequence of real, non-negative numbers, then if $\lim (b_n) = 0$ it must mean that $A$ is compact subset of $X$.
Take any sequence of sequences $(x_n)$, our goal is to construct a subsequence, $(x_{n_k})$ which will converge. Suppose for $n \geq N$ we have $|b_n|<\epsilon$ (by convergence of $(b_n)$). For the first $N-1$ points, however, I thought we could use Bolzanno-Weirstrass on each of the $N-1$ first terms (since $|x_n| \le b_n \forall n$) in the following way: apply Bolzano Weirstrass on all the first terms, then for the second terms apply it on the subsequence we got from the first terms, and so on... and claiming this subsequence will converge to $\{x_1,x_2,...,x_{N},...\}$ - edited, where $x_i$ is the limit of the $i_{th}$ subsequence.
However, the subsequence created could have a few cases where we end up with no terms at the end of this inductive argument. My teacher told me to use Cantor Diagonalization to avoid this, but I don't see how this would work.