Explicitly demonstrating non-compactness of closed unit ball in $l^{\infty}$

Question

It is true that the closed unit ball centered at 0 in $l^{\infty}$ (set of bounded real sequences) is not compact which can be shown in multiple ways by exploiting results such as "compactness implies sequential compactness".

I would like to demonstrate the above by explicitly constructing an open cover of $\bar{B}(0,1)$ with no finite subcover, showing directly from the definition of compactness the result stated. However, I am having difficulty constructing such an open cover. Can you please provide a hint rather than a full solution for how to proceed?

Thank you.

Answer 1

The idea is to find a contructive proof of Heine-Borel theorem implies Bolzano-Weierstrass Theorem and transpose it.

Let $\langle f_j \rangle_{j=0}^\infty \subseteq B^-$ be a sequence without a convergent subsequence. Then for every $x \in B^-$, there exists an open ball $B_x$ centred at $x$ which contains only finitely many $f_j$s. The open cover $(B_x)_{x \in B^-}$ has no finite subcover. (Otherwise $\langle f_j \rangle$ has a convergent subsequence)