I'm proving Theorem 3.18 in Brezis's book of Functional Analysis.
Assume that $E$ is a reflexive normed linear space and $E^\star$ its topological dual. Let $\left(x_{n}\right)$ be a bounded sequence in $E .$ Then there exists a subsequence $\left(x_{n_{k}}\right)$ that converges in the weak topology $\sigma\left(E, E^{\star}\right)$.
The author said that The proof of Theorem 3.18 requires a little excursion through separable spaces and will be given in Section 3.6. However, I have come across below proof which is very simple. I suspect I made some subtle mistakes. Could you have a check on my attempt?
Proof: Assume that $|x_n|$ is bounded by $M \in \mathbb R_+$. Then $(x_n / M)$ is a sequence in the closed unit ball which is compact in $\sigma\left(E, E^{\star}\right)$ by Kakutani's theorem. It follows that there exist $x\in E$ and a subsequence $(x_{n_k}/M)$ of $(x_n/M)$ such that $x_{n_k} / M \to x$ in $\sigma\left(E, E^{\star}\right)$ as $k \to \infty$. It follows that $x_{n_k} \to Mx$ in $\sigma\left(E, E^{\star}\right)$ as $k \to \infty$.
Your proof is only marginally incomplete I would say. As @MaoWao mentioned in their comment, compactness does not imply sequential compactness in general. However, as Brezis shows in Theorem 3.29, the unit ball is not only compact in $\sigma(E,E^\star)$ but metrisable and thus sequentially compact, as long $E^\star$ is separable (this is true because you can consider as $E$ the closure of space generated by your sequence which is clearly separable and also reflexive). It follows that thus you can find a subsequence and your proof works.