If $X$ is reflexive and the dual $X'$ contains a countable set which separates the points of $X$, show $X'$ is separable.
One way to prove this is using the fact that the unit ball $B_{X}$ is a weakly compact set whenever $X$ is reflexive, and since $X'$ has a countable set that separates the points of $X$, the weak topology on $B_{X}$ is metrizable, from which it follows that $X'$ is separable. But I can't use this idea because the notion of weak topology is discussed later in next chapter of my textbook, so I think using Hahn-Banach theorem is the right way to go but I have no idea how to start. Any hint is welcome.
Best Answer
Hint: It's enough to prove that $X$ is separable. Let, $S=\{f_n\}$ be countable collection of functionals which separates points of $X.$
Prove that, there exists $x_n$ such that, $X=\text{span} \{x_n\}\bigoplus \ker f_n.$ Call, $T= \{x_n\}$
Take, $D=\{\lambda x:\lambda\in\mathbb{Q}+i\mathbb{Q} ,x\in T\}$ and show that, $\bar{D}=X$
Edit(Elaborated): It's well known that, if $Y$ is a n.l.s such that $Y^*$ is separable then $Y$ is separable.
Since, we have shown above that $X$ is separable and $X$ being reflexive , so $X^{**}$ is also separable.
Therefore, $X^*$ is separable.