Question: > 
> 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?

Answer: 
Look at the straight line that is the set of all scalar multiples of one non-zero element of $\ell^2$. It's a closed subspace. If the whole space were compact, then that line would be compact. But that line is homeomorphic to the ordinary real line.
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.