[proofplan] We combine the [Banach-Alaoglu theorem](/theorems/212) (which gives weak* compactness of $B_R$) with the [weak* metrizability theorem](/theorems/495) (which makes $B_R$ a compact metrizable space when $X$ is separable). In compact metrizable spaces, compactness implies sequential compactness, which we prove via a diagonal extraction argument using total boundedness. [/proofplan] [step:Reduce to a compact metrizable space] Let $\|f_n\|_{X^*} \leq R$ for all $n$. By the [Banach-Alaoglu theorem](/theorems/212), $B_R = \{f \in X^* : \|f\|_{X^*} \leq R\}$ is compact in $\sigma(X^*, X)$. By the [Weak* Metrizability on Bounded Sets](/theorems/495) theorem, since $X$ is separable, the weak* topology on $B_R$ is metrizable. So $B_R$ is a compact metrizable space containing $\{f_n\}$. [/step] [step:Extract a convergent subsequence via sequential compactness of compact metric spaces] Let $(M, d)$ be a compact metric space. Since $M$ is compact and metrizable, it is totally bounded. Construct the subsequence by a diagonal argument: at stage $m$, cover $M$ by finitely many balls of radius $1/m$ and extract a subsequence lying in one ball. The diagonal subsequence $z_n := y_n^{(n)}$ is Cauchy: for $n, k \geq m$, both lie in a ball of radius $1/m$, so $d(z_n, z_k) \leq 2/m \to 0$. Since compact metric spaces are complete, $\{z_n\}$ converges. Applying this to $(B_R, d)$, the sequence $\{f_n\}$ has a subsequence $\{f_{n_k}\}$ converging in the weak* metric to some $f \in B_R \subseteq X^*$. In particular $\|f\|_{X^*} \leq R$ and $f_{n_k}(x) \to f(x)$ for every $x \in X$. [/step]