Theorem 19.5 in Munkres Let $\{X_\alpha\}$ be an indexed family of spaces; let $A_\alpha \subset X_ \alpha$ for each $\alpha$. If $\prod X_{\alpha}$ is given either the product or the box topology, then $$\prod \bar{A}_{\alpha} = \overline{\prod A_{\alpha}}$$ Proof. Let $x=(x_\alpha)$ be a point of $\prod \bar{A}_{\alpha}$; we show that $x \in \overline{\prod A_{\alpha}}$. Let $U=\prod U_{\alpha}$ be a basis element for either the box or product topology that contains $x$. Since $x_\alpha \in \bar{A}_{\alpha}$, we can choose a point $y_\alpha \in U_\alpha \cap A_\alpha$ for each $\alpha$. Then $y=(y_\alpha)$ belongs to both $U$ and $\prod A_{\alpha}$. Since $U$ is arbitrary, it follows that $x$ belongs to the closure of $\prod A_{\alpha}$.
I think that this part of the proof requires the Axiom of Choice because it says "choose a point $y_\alpha \in U_\alpha \cap A_\alpha$ for each $\alpha$". In addition, I know that "The Cartesian product of any nonempty family of nonempty sets is nonempty" is equivalent to the AC. So since we know $U_\alpha \cap A_\alpha$ is nonempty for each $\alpha$, we need AC to show that $U \cap \prod A_{\alpha} = \prod {U_{\alpha} \cap A_{\alpha}}$ is nonempty, proving that $x$ belongs to the closure of $\prod A_{\alpha}$.
Is there any way to prove this part of the theorem without using AC?
No. In fact, this theorem is equivalent to AC. To prove AC from it, suppose $(A_\alpha)$ is a family of nonempty sets which has no choice function. Let $*$ be a set that is not an element of any $A_\alpha$, and let $X_\alpha=A_\alpha\cup\{*\}$ with the indiscrete topology. Then $\overline{A_\alpha}=X_\alpha$ for each $\alpha$, so $\prod \overline{A}_\alpha=\prod X_\alpha$ and in particular is nonempty since it contains an element which is $*$ on every coordinate. On the other hand, $\prod A_\alpha$ is empty since $(A_\alpha)$ has no choice function, so $\overline{\prod A_\alpha}$ is also empty (in any topology). In particular, $\overline{\prod A_\alpha}\neq \prod \overline{A}_\alpha$.