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?