A weakly convergent sequence in a compact set, is strongly convegnet

Question

Let $E$ be a Banach space, and $K \subset E$, compact set for the strong topology.

And let $(x_n)_n$ converges for the weak topology $\sigma(E,E^*)$ to $x$.

Why $(x_n)_n$ converges for the strong topology ?

My idea :

Since $K$ is a compact set for the norm topology then $(x_n)_n$ has a convergent subsequence $(x_{n_k})_k$ for the norm topology to $x$ (Since $(x_{n_k})_k$ converges weakly to x).

How to prove that the sequence $(x_n)_n$ converges strongly to $x$ ?

I'm stuck in going from Since $(x_{n_k})_k$ converges weakly to x. then $(x_{n_k})_k$ to Since $(x_{n_k})_k$ converges weakly to x. then $(x_{n})_n$.

Answer 1

Here is another approach. Let $X$ denote $K$ equipped with the strong topology, and $Y$ denote $K$ equipped with the weak topology. Since the strong topology is finer than the weak topology, the identity map $X\to Y$ is continuous. But $X$ is compact and $Y$ is Hausdorff, so any continuous bijection $X\to Y$ is a homeomorphism. So the identity is a homeomorphism; that is, the weak topology and strong topology on $K$ are the same.

Answer 2

Suppose that $(x_n)$ doesn't converge in norm to $x$. Then there exists some $\varepsilon>0$ and a subsequence $(x_{n_k})$ such that $\|x_{n_k}-x\|\geq\varepsilon$ for all $k$. Since $(x_n)$ is contained in a compact set, we must have some sub-subsequence of $(y_\ell)$ of $(x_{n_k})$ which converges in norm. But this norm limit must be $x$, which contradicts $\|y_\ell-x\|\geq\varepsilon$ for all $\ell$.