Subset of $l^\infty$ compact or not I would be grateful if you can give me some hints for the following homework problem.

Let $C$ be a subset of $l^\infty$ (with uniform norm) such that $C = \left\{(x_n) \mid |x_n|\leq \frac1n \,\forall n\geq 1\right\}$
  Is $C$ a compact set or not?

Honestly, I am stuck. I tried to use sequentially compactness and attempted to construct different sequences which may serve as counter examples but all of them failed to do so.
I also thought that if $y_n$ is a constant sequence equal to $0$ and if we take the open ball $B(y_n,2)$ with radius $2$, does that count as a finite open cover of $C$?
Thanks in advance.
 A: Since C is normed, it is compact iff it is sequentially compact. Let $\{x_n\}\subset C$ be a sequence, then by the Cantor diagonal trick, it has a pointwise convergent subsequence $\{x'_n\}$, i.e. $x'_n(i)\to y(i)$ as $n\to\infty$ and every $i$. Clearly, $y\in C$, and for any $\epsilon>0$, choose $N>2/\epsilon$. Then, for all $i<N$, $|x'_n(i)-y(i)|<\epsilon$ for all $n$ large enough, while for $i\ge N$ we have $|x'_n(i)-y(i)|\le 1/N+1/N <\epsilon$. Thus $\{x'_n\}$ converges in norm, and thus C is sequentially compact.
A: First, we show that $C$ is closed. If $x=\{x_n\}\notin C$ then we can find $N$ such that $|x_N|>N^{-1}$, and the ball of cented $x$ and radius $\frac{|x_N|-N^{-1}}2$ is contained in the complement of $C$. 
Now, we use the result that a pre-compact subset in a complete space has a compact closure (hence here $C$ will be compact). 
Fix $\varepsilon>0$ and $N$ such that $N^{-1}\leq \varepsilon$. Since the product $\prod_{j=1}^{N-1}\left[-j^{-1},j^{-1}\right]$ is precompact, we can find an integer $p$ and $x^{1},\ldots,x^{p}\in \prod_{j=1}^{N-1}\left[-j^{-1},j^{-1}\right]$ such that for each $v\in \prod_{j=1}^{N-1}\left[-j^{-1},j^{-1}\right]$, we can find $1\leq i\leq p$ such that $\max_{1\leq j\leq p}|v_j-x_j^i|\leq \varepsilon.$ Now put for $1\leq j\leq p$, $y^j$ the sequence defined by $y^j_k=\begin{cases}x_k^j&\mbox{if }k<N\\\
0&\mbox{ otherwise.}\end{cases}$ 
For all $x\in C$, we can find $1\leq i\leq p$ such that $\lVert x-y_i\rVert\leq \varepsilon$, and you can conclude. 
A little more generally, you can show that if $\{a_n\}$ is a sequence which converges to $0$ then the set $C:=\left\{\{x_n\}\mid\forall n, |x_n|\leq a_n\right\}\subset \ell^{\infty}$ is compact. 
