Unit sphere compactness in a metric space Define $A$ as a nonempty set, $\mathcal{B}:=\{f: A \rightarrow \mathbb{R}: f(A) \text{is bounded} \} ,d_\infty:=\text{sup}\{|f(x)-g(x)|:x \in A\}$. 
For which $A$ is $\overline {B_1(0)} \subset \mathcal{B}$ compact?
Notes: ${B_1(0)}$ is the unit sphere. I already proved that $(\mathcal{B},d_\infty)$ is a metric space.
My thoughts: I investigated a bit on this and found some proofs that the unit sphere in a banach space is compact when the banach space is finite dimensional. But I don't want to use this here, I don't even know if $(\mathcal{B},d_\infty)$ is a banach space. All proofs we did who depend on the dimension of a metric space is that the intersections of compact subsets is compact and that the finite union of compact subsets is compact. Maybe we can use this?

The result you mention has a converse: the closed unit ball in a Banach space is compact if and only if the Banach space is finite-dimensional.  That should suggest to you what is true here.  
