 Solution

First suppose that $X$ is separable. Choose a norm-dense sequence $(x_n)$ in $B_X$. On every norm-bounded subset of $X$, a countable norm-dense subset $(f_n)$ of $B_{X^{\ast }}$ in the weak-star topology separates points, andmetrizes the weak topology. Indeed, convergence for all $f_n$, boundedness, and weak-star density imply convergence for every $f\in X^{\ast }$. Thus a weakly compact set $K$ is a compact metric space and hence is sequentially compact.

For general $X$, take a sequence $(x_n)$ in $K$ and letThe space $Y$ is separable and norm closed. The Hahn-Banach theorem shows both that $Y$ is weakly closed in $X$ and that its own weak topology is the subspace topology inherited from $X$. Hence $K\cap Y$ is weakly compact and, by the separable case, contains a weakly convergent subsequence of $(x_n)$. Its limit lies in $K\cap Y$. Therefore every weakly compact subset of a Banach space is weakly sequentially compact, which is one direction of the Eberlein-Smulian theorem.