$\newcommand{\span}{\operatorname{span}}$In here $S^0=\{f\in X':f|_S\equiv0\}$ the annihilator set of $S$. And for $T\subset X'$, we had ${}^0T=\{x\in X: \forall f\in T \text{ we had } f(x)=0\}$. Then, $${}^0(S^0)=\{x\in X: \forall f\in S^0, \text{ we had } f(x)=0\}.$$
The fact of ${}^0(S^0)\supset \overline{\span}(S)$ is possible showing, if im correct, leting $(y_n)\subset \span(S)$ such that $y_n\rightarrow x$, where
$$y_n=\sum^{k_n}_{i=1}\lambda_i y_i^n\quad\text{and}\quad f(x) = f\left(\lim_{n\rightarrow+\infty}y_n\right) = f\left(\lim_{n\rightarrow+\infty} \sum^{k_n}_{i=1}\lambda_i^n y_i^n\right) = \lim_{n\rightarrow+\infty} \sum^{k_n}_{i=1}\lambda^n_i f( y_i^n)=0,\forall f\in S^0.$$
Hence $x\in {}^0(S^{0})$.
But, for prove ${}^0(S^0)\subset \overline{\span}(S)$, we had one hint that is prove in other exercise $S^0=(\overline{\span}(S))^0$ and ${}^0T={}^0(\overline{\span}T)$. What implies in $${}^0(S^0)={}^0((\overline{\span}S)^0)\quad \text{and} \quad {}^0(S^0)={}^0(\overline{\span}(S^0))={}^0(\overline{\span}((\overline{\span}S)^0)).$$
In this point i dont know what do...
$\newcommand{\span}{\operatorname{span}}$Let $x\in {}^0((\overline{\span}S)^0)$, for all $f\in (\overline{\span}S)^0$, we had $f(x)=0$. Suppose that $x\notin \overline{\span}S$. How $\overline{\span}S$ is a subespace of $X$, we know of the Functional Analisys, that exists $f\in X'$ such that $f|_{\overline{\span}S}\equiv 0$, $\|f\|\leq 1$ and $f(x)=d(x,\overline{\span}S)$. Hence $x\in \overline{\span}S$.