Let $K$ be an algebraically closed field, $S \subset R = K[X_1,X_2,...,X_n]$ and $J = (S)$ the ideal generated by $S$. Let $V$ and $I$ denote the usual correspondences between $R$ and $K^n$
The Nullstellensatz states (among other things) that for any $J \subset R$:
$$ I(V(J)) = \sqrt{J}$$
Since $V(S) = V(J),$ we have $\sqrt{J} = \sqrt{S}.$
This proof can't be correct. Take $S = \{1\}$ then $J = R$ but
$$ \sqrt R = R $$ and
$$ \sqrt{\{ 1\}} = \{g \vert \exists n : g^n = 1 \} \neq R.$$
Where did I go wrong ?
As per request, I will write my comment as an answer. The problem here is that you have (mistakenly) taken the radical of the singleton set $\{1\}$. The singleton set $\{1\}$ is not an ideal. We would rather take radicals of ideals because then it can be shown quite easily that the radical of an ideal is also an ideal. Thus, we should take the radical of the ideal generated by $\textbf{1}$ which is denoted as $\sqrt{(1)}$.
So the correct formulation of your last line would be: $$\sqrt{(1)} = \{g \vert \exists n : g^n \in (1) = R \} = R.$$