Prove that the unit ball in $X$ is not compact 
Let $X$ bet the set of all sequences $\{a_n\}_{n=1}^\infty$ in
  $\mathbb R$ with $\lim_{n \to \infty} a_n = 0$. For any $\{a_n\},
 \{b_n\} \subset X$, we define a metric 
$$d(\{a_n\}, \{b_n\}) = \sup\{|a_n - b_n|, n = 1,2,\dots \}$$
Prove that the unit ball $\{\{a_n\}\in X: d(\{a_n\}, 0) = \sup_{n \in
 \mathbb N} |a_n| \le 1\}$ in $X$ is not compact

I think this unit ball is both bounded and closed in metric space, so it is compact. But the question asks me to prove that it is not compact. Please help.
 A: One of the more intuitive definitions/characterizations of compactness in a metric space $(X,d)$ is the following:
A set $K \subseteq X$ is compact if and only if every sequence $\{a_k\} \subseteq K$ has a convergent subsequence.
So in order to show that the unit ball in the space of all sequences is not compact, you have to find a sequence (of sequences) in the unit ball that does not contain a convergent subsequence.
A: If $K$ is a compact set in a Hausdorff space, the only closed discrete subsets of $K$ are the finite sets.
The infinitely many sequences $s_n$ where
$$
s_n(n)=1,\qquad \text{$s_n(m)=0$ for all $m\neq n$}
$$
form a closed, discrete, but infinite subset of the unit ball in the space of all convergent sequences, implying the latter is not compact.
A: For $S\subseteq \mathbb N$ let
$$ a^{(S)}_n=\begin{cases}1&\text{if $n\in S$}\\-1&\text{if $n\notin S$}\end{cases}$$
Then for any sequence $a\in X$ we have $d(a,a^{(S)})<2$ for $S=\{\,n\in \mathbb N\mid a_n>0\,\}$. Hence 
$$\tag1X=\bigcup_{S\subseteq \mathbb N} B(a^{(S)},2).$$
Consider a finite subcover $$\tag2B(a^{(S_1)},2)\cup \ldots \cup B(a^{(S_m)},2)$$
and let $T=\{\,n\in\mathbb N\mid 1\le n\le m, n\notin S_n\,\}$. Then $d(a^{(T)},a^{(S_n)})=2$ for $1\le n\le m$, so that $a^{(T)}$ is not covered by $(2)$.
Remark: Apparently $(1)$ doesn't even allow a countable subcover.
A: Consider 0= (0,0,0,……………) € l2
Consider the closed ball, B [0,1].
Clearly B [0,1] is a closed and bounded subset of l2 . (Any closed ball in a metric space is a closed set and diam B[0,1]
