$X^*$ is separable then $X$ is separable [Proof explanation] $X^*$ 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).
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 ($\clubsuit$).

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
 A: Take a bounded linear functional $f \in Y^{\ast}$ s.t. $f\vert_Y = 0$. Then choose some $u \in X \setminus Y$ with $\Vert u \Vert = 1$ (which exists since $Y$ is proper), define $g \in (Y \oplus \mathbb{C}u)^{\ast}$ via $g(u) = 1$ and $g\vert_Y = f$ and extend linearly. 
Then we have $\Vert g \Vert = 1$, since for every $x \in X$ with $\Vert x \Vert = 1$ we either have $g(x) = 1$ or $g(x) = 0$. Hence Hahn-Banach gives an extension $G \in X^{\ast}$ s.t. $\Vert G \Vert = \Vert g \Vert = 1$ and $G \vert_Y = g \vert_Y = f = 0$. 

The above uses the norm preserving version of Hahn-Banach. But really, you only need any extension of the above defined $g$. Then you can you may rescale, and since $g\vert_Y = 0$, the rescaled $g$ is also $0$ on $Y$. The sublinear functional used for this would be $x \mapsto \Vert g \Vert \cdot \Vert x \Vert$. 
A: Take $u\in X\setminus Y$ and define $f:Y\oplus \langle u\rangle\to \mathbb F$ by $f(y+cu)=c$. We can apply the Hahn-Banach theorem once we prove that $f$ is bounded. For this, use the fact that $Y$ is a closed subspace of $X$ so that there is a $\delta>0$ such that $\|y-u\|>\delta$ for all $y\in Y.$ Then, the claim follows from the calculation $\|y+cu\|=|c|\cdot \left \|\left(\frac{-1}{c}\right)y-u\right \|>c\delta.$
A: You only need the existence of a non-zero bounded linear functional on the quotient normed space $X/Y$, because composing it with the quotient map $X \rightarrow X/Y$ and then rescaling will give you the desired $f$.
