Bounded operator from a Hilbert space to $\ell^1$ is compact

Let $H$ be any Hilbert space. How can we prove that any bounded linear operator $T\colon H \to \ell^1$ is compact?

If we use the fact that the space $\ell^1$ has Schur property (norm and weak convergence is the same), then we need to show that for a sequence $(x_n) \subset H$ such that $||x_n|| \leq 1$ the sequence $(Tx_n) \subset \ell^1$ contains weakly convergent subsequence. But I do not know how I should do this.

What property of a Hilbert spaces I need to use?

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First note that you can assume that $H$ is separable (since the sequence $\{x_n\}$ is contained in a separable subspace of $H$. Then you are in a case covered by Pitt's theorem. If you have acccess, this article by Delpech may be helpful, see also this thread –  t.b. Nov 1 '11 at 0:17
@t.b.: Nice! Thanks. I don't know why I didn't thought about the reduction to separable case. –  xen Nov 1 '11 at 1:14
Oh, silly me -- I didn't notice that I linked to your old question. –  t.b. Nov 1 '11 at 1:24

First, we can assume $H$ is separable, since any sequence in $H$ is contained in a closed separable subspace.
Because $\ell^1$ has the Schur property, we just need to show that for every $x_n$ in the closed unit ball of $H$, $Tx_n$ has a weakly convergent subsequence. Since for every $f \in {\ell^1}^*$, $fT$ is a continuous linear functional on $H$, it's enough to show that $x_n$ has a weakly convergent subsequence. Thus it's enough to show that the closed unit ball of $H$ is weakly sequentially compact. Since $H$ is self-dual and separable, the sequential Banach-Alaoglu theorem tells us exactly what we want.