= Solution
\b[True.] Choose a norm-dense sequence $(x_n)$ in the unit ball of the separable <normed vector space> $E$. On the dual unit ball $B'$, define
$$
\boxed{d(\phi,\psi)=\sum_{n=1}^{\infty}2^{-n}|\phi(x_n)-\psi(x_n)|.}
$$
Each summand is at most $2^{1-n}$, so the series converges uniformly. Symmetry and the <triangle inequality> follow term by term. If $d(\phi,\psi)=0$, the two continuous functionals agree on the dense sequence, hence on the unit ball and then on all of $E$. Thus $d$ is a <metric>.
A relative weak-star neighbourhood controls finitely many evaluations. Conversely, to make $d$ small, first choose a tail whose sum is small and then control the finitely many evaluations at $x_1,\ldots,x_N$. This proves that each <metric> ball is weak-star open. For the other direction, let $x\in E$ and $\epsilon>0$. Approximate $x/\|x\|$ by $x_n$ when $x\ne0$. Since $\|\phi-\psi\|\leq2$ on $B'$,
$$
|\phi(x)-\psi(x)|
\leq\|x\|\,|\phi(x_n)-\psi(x_n)|+2\|x\|\,\|x/\|x\|-x_n\|.
$$
The second term can be made smaller than $\epsilon/2$ by choosing $n$; the first is smaller than $\epsilon/2$ whenever $d(\phi,\psi)$ is sufficiently small, because $|\phi(x_n)-\psi(x_n)|\leq2^n d(\phi,\psi)$. Every evaluation is therefore metric-continuous. These two inclusions show that $d$ induces exactly the <weak-star topology>, proving <weak-star metrizability of the dual ball> without a completeness assumption on $E$.
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