Proving a subspace is not compact Let $l^2 = \left\{ \left\{\alpha_n\right\}_{n\geq1} : \sum_1^\infty |\alpha_n|^2 < \infty\right\}$ with the usual metric.
Let $F =\left\{ \left\{\alpha_n\right\}_{n\geq1} : \sum_1^\infty |\alpha_n|^2 \le 1\right\} $
Show that $F$ is not compact.
The metric is $\sum |a_n - b_n|^2$
I was trying to prove $F$ is not compact by finding a continuous function from $F$ to $\mathbb{R}$ somehow involing the metric, and showing that the image is not compact, but cannot find any useful function.

One correction to start with: the usual metric is 
$$d(\alpha, \beta) = \sqrt{ \sum_{n=1}^{\infty}(\alpha_n - \beta_n)^2\ } $$
which extends the usual Euclidean metric to countably many coordinates. This is the standard Hilbert space $\ell^2$.
To see non-compactness consider the sequence basis sequence $e_n \in \ell^2$ defined by $(e_n)_m = 0$ for $n \neq m$, $1$ if $n=m$.
Check that all $e_n \in F$, $d(e_n, e_m) = \sqrt{2}$ for $n\neq m$, and conclude that $(e_n)_n$ has no convergent subsequence.
