$X^\*$ is separable then $X$ is separable  

**Proof:**

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)](https://rads.stackoverflow.com/amzn/click/com/0471317160).

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)](https://rads.stackoverflow.com/amzn/click/com/0471317160). 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 ($\clubsuit$).

> 
> I don't understand the part apply Hahn Banach Theorem. This part in special: 
> 
> 
> $\begin{align\*}
> f(y)=&\,0\quad\forall y\in Y,\\
> \|f\|=&\,1;
> \end{align\*}$
> 
> 
> 

Can someone clarify me that part in special? Thanks for answer!

The answer is of this post: [Showing $X^\*$ is separable implies $X$ is separable using the Riesz lemma](https://math.stackexchange.com/questions/1397622/showing-x-is-separable-implies-x-is-separable-using-the-riesz-lemma)