Problem: Let $\{X_\alpha\}_{\alpha \in J}$ be a family of topological spaces and $C_\alpha \subseteq X_\alpha$ a closed subset for all $\alpha \in J$. Then the product $\prod_{\alpha \in J} C_\alpha$ is closed in $X=\prod_{\alpha \in J} X_\alpha$ endowed with either the box or subspace topology.
To start I wanted to prove the statement when $X$ has the product topology. I had what I thought was a valid approach, but when I checked here I saw many answers use the fact that $\prod_{\alpha \in J} C_\alpha = \bigcap_{\alpha \in J} \pi_\alpha^\leftarrow(C_\alpha)$, which I hadn't thought to use. I see why that approach works, but I wanted to double check if my attempt made sense.
Attempt: Notice that $(x_\alpha) \in \prod_{\alpha \in J} X_\alpha - \prod_{\alpha \in J} C_\alpha$ means there is some index $\alpha_0 \in J$ for which $x_{\alpha_0} \notin C_{\alpha_0}$, note equivalently $x_{\alpha_0} \in X_{\alpha_0}-C_{\alpha_0}$ which is open, but there are no restrictions on the memberships of other components. For any index $\alpha_0 \in J$, let $\widehat{C_{\alpha_0}} = \prod_{\alpha \in J} Y_\alpha$ where $Y_\alpha = X_\alpha$ for all $\alpha \in J \setminus \{\alpha_0\}$ and $Y_\alpha = X_{\alpha_0}-C_{\alpha_0}$ for $\alpha= \alpha_0$, then $X- \prod_{\alpha \in J} C_\alpha = \bigcup_{\alpha \in J} \widehat{C_\alpha}$. It is clear that for some fixed $\alpha_0 \in J$ that $\widehat{C_{\alpha_0}}$ is a basic open set of either the box or product topology on $X$, and therefore $\prod_{\alpha \in J} C_\alpha$ is closed in $X$ as a union of such sets.
Let me know if this approach makes sense, and where it can be improved if it's clunky. Thanks in advance.
Here is an easier proof: $\prod_{\alpha\in J}C_\alpha$ is the intersection of $\prod_{\alpha\in J\backslash\{\beta\}}X_\alpha\times C_\beta$, which is closed. Arbitrary intersections of closed subsets are closed.