Given a set $X$ we say that a collection $\cal C$ in $X$ is totally closed with respect union or intersection if respectively the union or the intersection of any subcollection $\cal S$ of $\cal C$ is in $\cal C$.
Let be now $\mathcal U$ in $\mathscr P\big(\mathscr P(X)\big)$ and $\mathcal V$ in $\mathscr P\big(\mathscr P(Y)\big)$ two collections totally closed with respect union/intersection: so, I am investigating if the collection $$ \mathcal U*\mathcal V:=\{U\times V:U\in\mathcal U\wedge V\in\mathcal V\} $$ of $X\times Y$ is totally closed with respect union/intersection too but, unfortunately, I was not able to prove it or to find a counterexample so that I thought to put here a specific question where I ask to solve the issue. So, could someone help me, please?
ADDITIONAL EXPLANATIONS
Given a subcollection $\mathcal A$ of $\mathcal U\ast\mathcal V$ then for any $A$ in $\mathcal A$ there exist a unique $U_A$ in $\mathcal U$ and a unique $V_A$ in $\mathcal V$ such that the equality $$ A=U_A\times V_A $$ holds but unfortunately the equality $$ \bigcup_{A\in\mathcal A}(U_A\times V_A)=\left(\bigcup_{A\in\mathcal A}A\right)\times\left(\bigcup_{A\in\mathcal A}V_A\right) $$ does not holds so that if $\mathcal U$ is totally closed by union (if $\mathcal U$ was totally closed by intersection then replacing the union with the intersection above the equality does not holds too) then $\bigcup\mathcal A$ is not apparently in $\mathcal U\ast\mathcal V$ so that I suspect that this collection is not totally closed.
Finally working about a counterexample it seems that the product of Euclidean topology in $\mathbb R^2$ is a valid counterexample for totally closedness by union since union of two disjoint rectangles is not apparentely product of two open sets but I think that this must be proved and this it seems not trivial at first glance.