If $X$ is a Banach space and $X^*$ is separable, then $X$ is
separable.
Here, David Mitra mentions a proof using the Riesz lemma. However, I do not fully understand it.
You could also use Riesz' lemma:
Let $Y$ be a proper closed subspace of the normed space $X$ and
$0<\theta<1$. Then there is an $x_\theta$ of norm 1 for which $\Vert
x_\theta-y\Vert>\theta$ for all $y\in Y$.If $X$ were not seperable, you could use Hahn Banach to construct
uncountably many functionals $f_\alpha\in X^*$ with $\Vert
f_\alpha-f_\beta\Vert\ge \theta$ whenever $\alpha\ne\beta$.
If $X$ is not separable, why does that allow us to construct uncountably many separated functionals $f_\alpha$?
I see why this fact would imply the theorem. Taking a countable dense set and $\epsilon$ balls around each point in that set, we see that two elements of the uncountable set $\{ f_\alpha\}$ must lie in a single ball. But then they are at most $2\epsilon$ apart. Making $\epsilon$ small shows we can find two elements of the set less than $\theta$ apart, a contradiction.
Best Answer
Here is my favorite proof, which I think is simpler than both the one suggested by David C. Ullrich and the one I had given earlier, elaborating on David Mitra’s hint. It uses only the Hahn–Banach theorem, but not Riesz’s lemma. It is based on the hint presented in Exercise 5.25, Folland (1999, p. 160).
If $X^*$ is separable, let $\{f_n\}_{n\in\mathbb N}$ be a countable dense subset of it. By the definition of the operator norm $$\|f_n\|\equiv\sup_{\substack{x\in X\\\|x\|\leq 1}}|f_n(x)|,$$ it is possible, for each $n\in\mathbb N$, to choose some $x_n\in X$ such that $\|x_n\|\leq 1$ and $$|f_n(x_n)|\geq\frac{1}{2}\|f_n\|\tag{$\clubsuit$}$$ (if $f_n=0$, then simply choose $x_n=0$; if $\|f_n\|>0$, use the definition of the supremum).
Let $C\equiv\{x_1,x_2,\ldots\}$. I claim that $\operatorname{span} C$ is dense, which implies that $X$ is separable (see the last claim in my previous post). To see this, suppose, for the sake of contradiction, that $\operatorname{span} C$ is not dense; then $Y\equiv\overline{\operatorname{span} C}$ is a proper closed subspace. By the Hahn–Banach theorem, it is possible to choose $f\in X^*$ such that \begin{align*} f(y)=&\,0\quad\forall y\in Y,\\ \|f\|=&\,1; \end{align*} see again Theorem 5.8(a) in Folland (1999, p. 159). Since $\{f_n\}_{n\in\mathbb N}$ is dense in $X^*$, there exists some $n\in\mathbb N$ such that $\|f_n-f\|< 1/3$. But then \begin{align*} |f_n(x_n)|=|f_n(x_n)-\underbrace{f(x_n)}_{=0}|\leq\|f_n-f\|<\frac{1}{3},\tag{$\diamondsuit$} \end{align*} whereas \begin{align*} 1=\|f\|\leq\|f-f_n\|+\|f_n\|<\frac{1}{3}+\|f_n\|, \end{align*} so that $\|f_n\|>2/3$. Putting this into ($\diamondsuit$), $$|f_n(x_n)|<\frac{1}{3}<\frac{1}{2}\|f_n\|,$$ which contradicts ($\clubsuit$).