$l^2$ is not compact 
Prove that the space $l^2$ (of real series $a_n$ such that $\sum_{i=1}^{\infty}a_i^2$ converges) is not compact.

I want to use the open cover $\{\sum_{i=1}^{\infty}a_i^2<n\mid n\in\mathbb{Z}^+\}$ and show that it has no finite subcover. To do that, I must prove that for any $n$, the set $\{\sum_{i=1}^{\infty}a_i^2<n\}$ is an open set. 
So let $\sum_{i=1}^{\infty}a_i^2<n$. Suppose it equals $n-\alpha$. I must find $\epsilon$ such that for any series $\{b_i\}\in l^2$ with $\sum_{i=1}^{\infty}(a_i-b_i)^2<\epsilon$, we have $\sum_{i=1}^{\infty}b_i^2<n$.
But $(a_i-b_i)^2$ has the term $-2a_ib_i$. How should I deal with that?

There're two easy ways to prove that $\ell_2$ is not compact. First, is to say that its dimension is infinite, hence closed unitary ball is not compact, hence the space itself is not compact.
Another way to see this is to find a bounded sequence which doesn't have a convergent subsequence; as a matter of fact, a sequence of basis vectors doesn't converge in norm (nor does any its subsequence); it's easy to see by checking Cauchy criterion.
