If $X^\ast $ is separable $\Longrightarrow$ $S_{X^\ast}$ is also separable Let $X$ be a Banach space such that $X$* (Dual space of $X$) is separable 
How can we prove that $S_{X^\ast}$ (Unit sphere of $X$*) is also separable
Any hints would be appreciated.
 A: Note: here we consider the question for $X^*$, and $S_{X^*}$, equipped with the topology induced by the norm. See this thread for the same question with the weak* topology.
This is a consequence of a much more general fact which has nothing to do with duals and Banach spaces.

Fact: if $(X,d)$ is a separable metric space, then for any subset $Y\subseteq X$, the subspace $(Y,d)$ is separable.

Note: as pointed out by Nate Eldredge, another approach is to observe that a subspace of a second countable space is obviously second countable. So it only remains to use or prove that a metric space is separable if and only it is second countable.
Proof: let $S=\{x_n\,;\,n\geq 1\}$ a dense subset of $X$. For every $n\geq 1$, consider the distance $d(x_n,Y)=\inf\{d(x_n,y)\,;\,y\in Y\}$. By definition of an infimum=greatest lower bound, for every $k\geq 1$, $d(x_n,Y)+\frac{1}{k}$ is not a lower bound of $\{d(x_n,y)\,;\,y\in Y\}$.  So there exists $y_n^k\in Y$ such that $d(x_n,y_n^k)< d(x_n,Y)+\frac{1}{k}$. 
Since $T=\{y_n^k\,;\,n\geq 1, k\geq 1\}$ is a countable union of countable sets, it is countable. We will now show that it is dense in $Y$.
Take $y\in Y$ and $\epsilon>0$. By density of $S$ in $X$, there exists $n\geq 1$ such that $d(x_n,y)<\frac{\epsilon}{2}$. In particular, $d(x_n,Y)\leq d(x_n,y)<\frac{\epsilon}{2}$. So for $k$ large enough, precisely: $k>\left(\frac{\epsilon}{2}-d(x_n,Y)\right)^{-1}$, we have $d(x_n,Y)+\frac{1}{k}<\frac{\epsilon}{2}$. Then for this $n$ and such a $k$, we get
$$
d(y_n^k,y)\leq d(y_n^k,x_n)+d(x_n,y)<d(x_n,Y)+\frac{1}{k}+\frac{\epsilon}{2}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.
$$
So there are elements of $T$ arbitrarily close to $y$ for every $y\in Y$. That is $T$ is dense in $Y$. QED.
Remark: what really is about Banach spaces $X$ such that $X^*$ is separable is that $X$ is separable. And it follows that the unit sphere of $X$, $S_X$, is separable as well.
