= 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 $B_{X^*}$ in the weak-star topology separates points, and
$$
d(x,y)=\sum_{n=1}^\infty2^{-n}
\frac{|f_n(x-y)|}{1+|f_n(x-y)|}
$$
metrizes the <weak topology>. Indeed, convergence for all $f_n$, boundedness, and weak-star density imply convergence for every $f\in X^*$. 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 let
$$
Y=\overline{\operatorname{span}}\{x_n:n\geq1\}.
$$
The 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 set>[weakly sequentially compact], which is one direction of the <Eberlein-Smulian theorem>.
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