I have to show that, $S\subseteq I(V(S))$ for $S\subseteq K[x_1,\dots,x_n]=:R$ where $I(X)=\{f\in R:f(a)=0,\forall a \in X\}$ and $V(Y)=\{p\in K^n:f(p)=0,\forall f \in Y\}$
I have already shown that $X\subseteq V(I(X))\tag1$
can I not define that $S=I(\tilde S)$, for some $\tilde S\subset K^n$ and then using $(1):$ $\tilde S\subseteq V(I(\tilde S))$ and since $I$ is inclusion reversing $S=I(\tilde S)\supseteq I(V(I(\tilde S)))=I(V(S))$ yielding a contradiction ?
or whenever $S$ is of the form $I(\tilde S)$ then the equality must hold ?
Hint: Take an arbitrary element of $a \in V(S)$ and an element $f \in S$. Now show that $f(a)=0$. Since $a$ was arbitrary you get $f \in I(V(S))$.
Take as example what Daniel said (there are ideals, which are not vanishing ideals (closed in the Zariski-topology)). On the other hand you are right, whenever $S$ is of the form $I(M)$ for some set $M \subseteq K^n$ then equality holds, meaning in the Zariski-topology that closure of a closed set is already the closed set. (If you don't know the Zariski-topology, just ignore this. You will probably learn soon, what it is).