I've the following result: Given any family of set $\{S_k\}_{k\in K}$, then $$\tag{*} \operatorname{cl}\left(\prod_{k\in K}S_k\right) =\prod_{k\in K}\operatorname{cl}(S_k), $$ and I'm looking for $$\tag{a} \operatorname{int}\left(\prod_{k\in K}S_k\right) =\prod_{k\in K}\operatorname{int}(S_k). $$
Further I use, with some freedom in the notation, $$\tag{b} \left(\prod_{k\in K} S_k\right)^C =\bigcup_{k\in K} \left(S_k{}^C\times\prod_{j\in K-k} X_j\right) $$ and $$\tag{c} \bigcup_{k\in K}\operatorname{cl}(S_k)\subseteq \operatorname{cl}\left(\bigcup_{k\in K}S_k\right), $$ and $$\tag{d} \operatorname{int}(S)=\big(\operatorname{cl}(S^C)\big)^C $$ Then $$ \begin{align} \operatorname{int}\left(\prod_{k\in K}S_k\right) &=^{(e)} \left(\operatorname{cl}\left(\left(\prod_{k\in K}S_k\right)^C\right)\right)^C \\ &=^{(f)} \left(\operatorname{cl}\left(\bigcup\left(S_k^C\times\prod_{j\in K-k}X_j\right)\right)\right)^C \\ &=^{(g)} \left(\bigcup\operatorname{cl}\left(S_k^C\times\prod_{j\in K-k}X_j\right)\right)^C \\ &=^{(h)} \left(\bigcup\left(\operatorname{cl}(S_k^C)\times\prod_{j\in K-k}X_j\right)\right)^C \\ &=^{(i)} \bigcap\left(\operatorname{cl}(S_k^C)\times\prod_{j\in K-k}X_j\right)^C \\ &=^{(j)} \bigcap\left(\left(\operatorname{cl}(S_k^C)^C\times\prod_{j\in K-k}X_j\right)\cup \left(\operatorname{cl}(S_k^C)\times X_j^C\times \prod_{\ell\in K-k-j}X_\ell\right)\right) \\ &=^{(k)} \bigcap\left(\operatorname{cl}(S_k^C)^C\times\prod_{j\in K-k}X_j\right) \\ &=^{(l)} \prod_{k\in K}\operatorname{cl}(S_k^C)^C\\ &=^{(m)} \prod_{k\in K}\operatorname{int}(S_k), \end{align} $$ where in (e) we applied (d); in (f) we applied (b); in (g) we applied (c) modified with the equality; in (h) we applied (*); in (i) we applied De Morgan; in (j) we applied (b) again; in (k) we applied $X_j^C=\varnothing$; in (l) we applied a not given formula; and in (m) we applied (d) again.
Question: Is the claim (a) correct? Are the used result (b) and (c) modified with the equality (I need the equality) correct?
Claim a) is wrong as soon as you do not have finitely many sets, because a product of open subsets is then not open in the product set with the product topology. There are, however, other topologies one can use on a product space, such as the box topology. https://en.m.wikipedia.org/wiki/Box_topology
The issue in your proof is at step (g), because closure does not commute with reunion in general — only with finite reunion.