$R/(IJ)$ is reduced $\Rightarrow IJ = I \cap J$ for ideals $I,J$ of a commutative ring $R$

The proof is correct and the statement doesn't indeed require $R$ is commutative.

What fails in the noncommutative case is that the set of nilpotent elements may not be an ideal, but it is not a problem here.

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You may want to note that a ring has no (nonzero) nilpotent elements if and only if $a^2=0$ implies $a=0$, so the use of just $a^2$ in your proof is not surprising.

Your question is a special case of this theorem.

Theorem: Let $R$ be a reduced ring, and let $I$ and $J$ be two ideals of $R$. If $IJ=0$, then $I\cap J=0$.

Proof: As your proof, since $(I\cap J)^2\subseteq IJ=0$, we have $(I\cap J)^2=0$. Thus, $I\cap J=0$.