For the inclusion $\overline S \subseteq VI(S)$, note that it suffices to show that $S \subseteq VI(S)$ and that $VI(S)$ is closed. By definition of the Zariski topology, $VI(S)$ is closed, so we only have to show $S \subseteq VI(S)$.
Let $P \in S$. If $f$ is a polynomial in $I(S)$ it vanishes at every point of $S$, in particular $f(P) = 0$. This shows the first inclusion.
For the converse, remember that $I$ and $V$ reverse inclusions, i.e. if $S_1 \subseteq S_2$ are subsets of $X$ then $I(S_2) \subseteq I(S_1)$, and if $A_1 \subseteq A_2$ are subsets of $K[X_1,\ldots,X_n]$ then $V(A_2) \subseteq V(A_1)$.
Since $\overline S$ is closed, there is some ideal $A \subseteq K[X_1,\ldots,X_n]$ s.t. $\overline S = V(A)$. Applying $I$ to the inclusion $S \subseteq V(A)$ yields $IV(A) \subseteq I(S)$. It is clear that $A \subseteq IV(A)$, hence we have $A \subseteq I(S)$. Applying $V$, we finally get $VI(S) \subseteq V(A) = \overline S$.