Separable dual implies space is weakly metrisable

For $E$ a Banach space and $E'$ (the continuous dual) a separable space, show that the closed unit ball $\overline{B}_{E} \subset E$ is weakly metrisable.

My work so far: As $E'$ is separable, we know that the closed unit ball $\overline{B}_{E''} \subset E''$ in the weak-$*$ topology is metrisable. Then Goldstine's lemma says that $\Phi(\overline{B}_E)$ is weak-$*$ dense in $\overline{B}_{E''}$ and we know $\Phi$ is an isometry, so $\overline{B}_E$ is metrisable. Finally, $\Phi^{-1}: (E'', \text{weak-}*) \to (E, \text{weak})$ is continuous (proved earlier), so the induced topology on $\overline{B}_E$ agrees with the weak topology.

($\Phi$ refers to the evaluation map, $E \to E''$ defined as $f \mapsto \varphi_f$ with $\varphi_f(F) = F(f)$ for $F \in E'$).

My question: Is this correct? It doesn't seem to me like I used the full power of Goldstine's lemma. Also, is there a more direct proof using $E'$ separable $\Rightarrow E$ separable?

Your logic is correct, but I don't think you even need Goldstine's lemma here. You already know that weak* topology is metrizable on bounded subsets of $E''$. The restriction of this topology to $E\subset E''$ agrees with the weak topology on $E$, by definition (both come from the same set of linear functionals). And any subspace of a metrizable topological space is metrizable, it need not be dense.
That said, a typical direct proof of the stated result is to pick a countable dense subset $\{\phi_n\}$ of the unit ball of $E'$ and verify that the metric $$d(x, y) = \sum_{n=1}^\infty 2^{-n} |\phi_n(x-y)|$$ metrizes the weak topology on the unit ball of $E$.
• Thank you for the reply! I felt like a direct proof as you say would be slightly wasteful, since that method was used to show the quoted result that the closed unit ball in $E''$ is weak-$*$ metrisable, so something higher level should be doable. I see how Goldstine isn't necessary though! – B. Mehta Apr 10 '18 at 1:54