 Solution

True in the specified topology. Let $D\subseteq B^{\prime }$ be the countable dense set obtained in part (ii). Scaling by a positive integer is a homeomorphism for the weak-star topology, so $nD$ is dense in $n{B}^{\prime }$. Sincethe countable set ${\bigcup }_{n}nD$ is dense in $E^{\prime }$: a nonempty open set contains a point of some $n{B}^{\prime }$, so its relative open intersection with that ball meets $nD$. This proves weak-star separability of the entire dual.

ThusThere is no assertion of norm separability here. For example, the dual of the separable Banach space ${\ell }^1$ is ${\ell }^{\infty }$, whose binary sequences form an uncountable set with pairwise norm distance one. That example is nevertheless separable in its weak-star topology.