How do I show that a closed subset $K\subset \cal l^p$ is compact? 
Let $1\leq p<\infty$. Let $K\subset \cal l^p$ be a closed subset. Assume that $\sup\{||x||_p:~x\in K\}<\infty$. I need to prove that $K$ is compact.

I think that with compact in this case we mean either sequentially compact or cover compact, but since both are equivalent we can chose which one we want to show.
My idea was to prove that $K$ is sequentially compact since it seems to work better.
Therefore let me pick a sequence $x^{(n)}\in K$, then by assumption $$||x^{(n)}||_p=\left(\sum_{k=1}^\infty |x^{(n)}_k|^p\right)^{\frac{1}{p}}<\infty$$
Now what I need to show is that there exists a convergent subsequence of $x^{(n)}$. Using that $K$ is closed would imply that the limit lies in $K$ and hence $K$ is sequentially compact.
My problem is now to find this convergent subsequence. Could maybe someone help me how to prove this?
Thanks a lot.
 A: The condition stated is not enough to show compactness, the example of $\{ e_k \}_k$ is an example that has no convergent subsequences.
If one assumes some uniformity as in for all $\epsilon>0$ there is some $N$ such that for any $x \in K$ we have
$\sum_{n>N} |x(k)|^p < \epsilon$ then the result it true.
A pedantic note about notation: If $x \in l^p$, I write $x=(x(1), x(2),...)$.
One straightforward proof is using the complete and totally bounded characterisation of compactness.
Since $K$ is closed it is complete, it remains to show that it is totally bounded.
Choose $\epsilon>0$ and $N$ such that the above condition holds.
The set $K_N=\{ (x(1),...,x(N)) | x \in K \} \subset \mathbb{R}^N$ is compact and so there is a finite $\epsilon$-net $y_1,...,y_{m_N}$ such that
$B(y_k, \epsilon)$ is a cover for $K_N$ ( with the $N$ dimensional $p$-norm).
Then $Y = \{ (y(0),...,y(N),0,0,...) | y \in Y_N \}$ is a $2 \epsilon$ net for $K$.
Addendum:
A collection of points $y_1,...,y_n \in Y$ is called an $\epsilon$-net for $Y$ iff $B(y_k, \epsilon)$ is a cover of $Y$.
A set if totally bounded iff for all $\epsilon>0$ there is an $\epsilon$-net.
