A closed ball in $l^{\infty}$ is not compact Definition of compact in Real Analysis, Carothers, 1ed said that:

In Example 8.1 (c), he claimed that closed ball $\{x: \|x\| \leq 1\}$ in $l^{\infty}$ is not compact. Why?
 A: compact implies limit point compact.
Hint : consider $\{(1, 0 , 0 , 0, \cdots) , \hspace{2mm} (0,1,0,0,0 \cdots), \hspace{2mm} (0,0,1,0,0,0, \cdots),  \cdots \}$. prove that it has no limit point. Actually you can prove stronger result that, they are discrete. But the only discrete subsets of a compact sets are finite.
A: If the ball was compact, so would its closed subset $\{e_n \ : \ n \in \mathbb{N}\}$. It is a general fact in topology (and not too difficult to prove) that a closed subset of a compact Hausdorff space is compact (the metric spaces automatically Hausdorff). Since the latter is not compact, the former is also not compact.
The reason why $\{e_n \ : \ n \in \mathbb{N}\}$ is not compact is that it is discrete (open unit balls at $e_n$ with radii $1/2$ are mutually disjoint), and countable, and hence has sequences without accumulation points. In fact, $e_1,e_2,\dots$ is one such sequence. You may also just start by pointing out that this series lies in the ball, and has no convergent subsequence, therefore the ball is not compact.
In general, you might be interested to learn that a ball in a Banach space (think: a vector space with well-behaved notion of distance) of infinite dimension always is non-compact.
A: Consider the sequence:
$$(1, 0 , 0 , 0, \cdots) , \hspace{2mm} (0,1,0,0,0 \cdots), \hspace{2mm} (0,0,1,0,0,0, \cdots),  \cdots$$
