$l^2$ 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?

I prefer the following approach more so than the ones already mentioned, perhaps it's useful for somebody else at some point:
Let $f\colon \ell^2\to \mathbb{R}$ be defined by $(x_n)_n\mapsto x_1$. Then $f$ is not bounded and thus $\ell^2$ can't be compact.
