Let $E$ be a Banach space and $S\subset E^{\ast}$ a subspace which is $w^{\ast}$-closed. Prove that there exists a subspace $S_0\subset E$ such that $S=\text{ann }(S_0)$, where $\text{ann }(X):=\left \{\varphi \in E^{\ast}:\varphi (X)=\left \{0\right \}\right \}$.
I was trying to solve this exercise by proving that for every subspace $S\subset E^{\ast}$, $\text{ann }(^{\perp }S)=\overline{S}^{w^{\ast}}$, where $\overline{S}^{w^{\ast}}$ is the $w^{\ast}$-closure of $S$.
It is straightforward to prove that $\overline{S}^{w^{\ast}}\subseteq \text{ann }(^{\perp }S)$. How about the other inclusion?
I found the following article: https://math.la.asu.edu/~quigg/teach/courses/578/2008/notes/adjoints.pdf
If you look at Proposition $8$, it is (I think) what we are trying to prove. But I could not understand the proof: Suppose we have $f\in \text{ann }(^{\perp }S)\setminus \overline{S}^{w^{\ast}}$. Then the author says that we can make use of Hahn-Banach theorem in order to obtain a $w^{\ast}$-continuous operator $\psi$ which annihilates $\overline{S}^{w^{\ast}}$ but $\psi (f)\neq 0$. Why is that true? I mean, it is true that, since $\overline{S}\subset \overline{S}^{w^{\ast}}$, then $\text{ann }(^{\perp }S)\setminus \overline{S}^{w^{\ast}}\subset \text{ann }(^{\perp }S)\setminus \overline{S}$, so we obtain a $\psi \in E^{\ast\ast}$ which satisfies the condition, but I cannot see why we can take that $\psi$ to be $w^{\ast}$-continuous.
Moreover, where is the Banach condition used?
