Artificial intelligent assistant

Is the product of pro-$\mathcal C$ groups also a pro-$\mathcal C$ group? Let $\mathcal C$ be a class of finite groups which is closed for subgroups and direct products and call a topological group $G$ a pro-$\mathcal C$ group if it is an inverse limit of $\mathcal C$-groups. I'm trying to show that a product $\prod_{i\in I}G_i$ of pro-$\mathcal C$-groups is a pro-$\mathcal C$ group and I'd be grateful for some help. I know that a pro-$\mathcal C$ group is isomorphic to a closed subgroup of a product of $\mathcal C$-groups- but I'm not sure if an arbitrary product of closed topological spaces is closed in the product topology.

Let $G=\prod_{i\in I}G_i$; in $G$ consider all normal subgroups $N$ which are of the following form: $$(\prod_{i\
e i_1,\dots,i_n}G_i)\times N_{i_1}\times\cdots N_{i_n}$$ where $\\{i_1,\dots,i_n\\}$ runs through all finite subsets of $I$ and the $N_{i_k}$ is a closed normal subgroup of $G_{i_k}$ such that the quotient $G_{i_k}/N_{i_k}$ is in $\mathcal C$. Then $G/N$ belongs to $\mathcal C$ for every such $N$, the latter follows from that fact that $G/N$ is a subdirect product of the $G_{i_k}/N_{i_k}$ and the assumption that $\mathcal C$ is closed under subgroups and (finite) direct products, and hence is also closed under (finite) subdirect products. The set of all such normal subgroups of $G$ is closed under finite intersection, so forms a directed set and the projective limit running over all such $G/N$ gives $G$.

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