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I'm doing Ex 3.7 in Brezis's book of Functional Analysis. Could you have a check on my attempt?

Let $E$ be a Banach space and let $K \subset E$ be a subset of $E$ that is compact in the strong topology. Let $\left(x_{n}\right)$ be a sequence in $K$ such that $x_{n} \rightharpoonup x$ weakly in $\sigma\left(E, E^{\star}\right)$. Prove that $x_{n} \rightarrow x$ strongly.

I posted my proof separately so that I can accept my own answer and thus remove my question from unanswered list. If other people post answers, I will happily accept theirs.

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    A compact topological space does not admit a strictly weaker Hausdorff topology. Hence, the topological spaces $(K,$strong) and $(K,$weak) are identical and thus have the same convergent sequences (and nets). – Jochen Feb 22 '22 at 13:32

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Lemma: Let $X$ be a topological space, $x\in E$, and $(x_d)_{d\in D}$ a net in $E$. Then $x_d \to a$ if and only if every subnet of $(x_d)_{d\in D}$ has a further subnet which converges to $a$. [A proof is given here]

Let $(n_k)_k$ be a subsequence of $\mathbb N$. Then by strong compactness of $K$, there is a subsequence $(n_{k_t})_t$ of $(n_k)_k$ and $a\in K$ such that $x_{n_{k_t}} \to a$ as $t \to \infty$. Because strong convergence implies weak convergence, $x_{n_{k_t}} \rightharpoonup a$ as $t \to \infty$. It follows from $x_{n} \rightharpoonup x$ that $x_{n_{k_t}} \rightharpoonup x$ as $t \to \infty$. Because weak topology is Hausdorff, $a=x$. By our Lemma, $x_n \to a$.

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