= Solution
Choose a norm-dense sequence $(f_n)$ in the unit sphere of the separable dual $X^*$. For every $n$, choose $x_n\in B_X$ with $|f_n(x_n)|>1/2$. If a functional $f\in X^*$ vanished on the closed linear span of the $x_n$, normalize it and choose $f_n$ with $\|f-f_n\|<1/4$; then $|f(x_n)|>1/4$, a contradiction. The <Hahn-Banach theorem> therefore shows that the span of $(x_n)$ is dense, so $X$ is separable.
Choose a norm-dense sequence $(u_n)$ in $B_X$. On $B_{X^*}$ define
$$
d_*(f,g)=\sum_{n=1}^\infty2^{-n}
\frac{|(f-g)(u_n)|}{1+|(f-g)(u_n)|}.
$$
Weak-star convergence implies convergence in this metric. Conversely, metric convergence gives convergence on the dense set $(u_n)$, and the uniform norm bound on the dual ball extends it to every $u\in X$. Thus $d_*$ metrizes the weak-star topology, proving <weak-star metrizability of the dual ball>.
Similarly, for a norm-dense sequence $(f_n)$ in $B_{X^*}$,
$$
d(x,y)=\sum_{n=1}^\infty2^{-n}
\frac{|f_n(x-y)|}{1+|f_n(x-y)|}
$$
metrizes the weak topology on $B_X$. Indeed, convergence against the dense functionals extends to every $f\in X^*$ because $x-y$ remains norm bounded.
Solved by gpt-5.6-sol high.
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