Weaky compact subset of Banach space with separable predual Let $X$ be a Banach space and $S\subseteq X$ be a subspace such that the unit sphere of $S$ is weakly compact. If $Y^*=X$ for some separable Banach space $Y,$ is it true that $S$ is separable?
 A: Assume that Samya Kumar Ray meant that the unit ball of $S$ is weakly compact. If $S$ is really a sphere, the fact that the answer is "Yes" follows from the argument outlined by Bill Johnson (and the additional assumptions are not needed).
The answer is "yes" anyway. Possibly a bit more straightforward argument than the one of Dirk Werner: the weak$^*$ topology on the unit ball of $S$ is weaker than the weak topology, so by the well-known result on compacts this implies that on the unit ball of $S$ weak$^*$ topology is the same as the weak topology. It is well-known and easy to show that bounded sets in the dual to a separable Banach space are weak$^*$ separable. Thus the unit ball of $S$ is weak separable. It remains to prove the easy statement that Banach spaces whose balls are weakly separable are separable.    
A: I found a proof using a theorem in "Topcis In Banach Space Theory" by Albiac and Kalton.
Theorem: Let $X$ be a Banach space. The Dual $X^*$ contains a sequence $(x_n^*)$ having the property $X_n^*(x)=0$ for all $n\geq 1$ implies $x=0.$ Then any weakly compact subset of $X$ is metrizable in weak topology.
Now to the original question take a sequence $(y_n)\subseteq Y\subseteq Y^{**}=X^*$ such that $(y_n)$ is dense in $Y$. Clearly  $(y_n)$ is total. The result follows applying the above theorem.
