Compact operator maps weakly convergent sequences into strongly convergent sequences I found the following property of compact operators in a proof, and I can't prove it.

Prove that if $T \in \mathcal{L}(E,F)$ is compact, and if $u_n \rightharpoonup u$ (the sequence converges weakly to $u$ in $\sigma(E,E^*)$) then $Tu_n \to Tu$ strongly in $F$.

I was able to prove that $Tu_n$ has a convergent subsequence ($u_n \rightharpoonup u$ implies that $(u_n)$ is bounded in $E$. Then, because $T$ is compact it must follow that $(Tu_n)$ must contain some strong convergent subsequence), but didn't manage to finalize the proof.
Any reference or hints are welcome.
 A: Make use of the following topological lemma. 
Lemma Let $X$ be a topological space and $\mathbf{x}=(x_n)_{n\in \mathbb{N}}$ be a sequence of elements of $X$. If every subsequence of $\mathbf{x}$ contains a subsequence convergent to $x$ then $x_n \to x$. 
A: This is also true by the following reason:
Since you've already proved that there is a strongly convergent subsequence, let's say $ Tu_{n_k} \to u^* $ for $ k \to \infty $. Then by the weak convergence of $ u_n \rightharpoonup u $ you get immediately that $ Tu_n \rightharpoonup Tu $. 
Now since strong convergence implies weak convergence and from the uniqueness of the limit of a weak convergent sequence it must be true that
$ u^*= Tu $.
Therefore $ Tu $ is a limit point of the sequence $ (Tu_n)_{n \in \mathbb{N}} $.
Now there's just one thing left to prove your statement
Claim: There's no other limit point, hence it must be the limit.
Proof: Suppose there's another limit point $ z $ of the sequence $ (Tu_n)_{n \in \mathbb{N}} $. Again there must be a subsequence $ (Tu_{n_m})_{m \in \mathbb{N}} $ converging to $ z $. Hence this subsequence $ u_{n_m} \rightharpoonup u$.
Last step, use the same argumentation as above to conclude that $ z = Tu = u^*$.
Therefore $ Tu_n \to Tu $ as $ n\to \infty $.
To be precise, at this point you know that
$ Tu_{n_k} \to Tu $ and that the limit $ Tu $ is the only limit point of $ (Tu_n)_{n\in \mathbb{N}} $. Use now a contradiction argument to prove that $ Tu_n\to Tu $.
