# closed bounded subset in metric space not compact

Let $\ell^{\infty}$ be the space of bounded sequences of real numbers, endowed with the norm $\|\mathbf x\|_\infty=\sup_{n\in N}|x_n|$, where $\mathbf x=(x_n)_{n\in\Bbb N}$.

Prove that the closed unit ball of $\ell^{\infty}, B'(\mathbf 0,1)={x \in \ell^{\infty} ; \|\mathbf x\|_{\infty} \le 1}$}, is not compact.

So basically I think I want to show that the implication that a closed and bounded subset of a metric space is compact is not necessarily true. In order to do this I think I will show it is not sequentially compact as the question tells me I am allowed to use the equivalence between compactness and sequential compactness in the setting of metric spaces. However, I am unsure if this approach is correct and do not know what sequence to use to show that the sequential compactness fails. Any help, hints or explanations are appreciated!

Hint: Show that there is a sequence of points $e_i$ on the unit sphere such that for $i\neq j$, $\|e_i-e_j\|_\infty=1$. Conclude that this sequence does not have a convergent subsequence, and therefore the unit ball is not compact.