So I know that if $A$ is a subset of a set $X$ and if $X$ is a subset of a set $Y$ then the equalities $$ \begin{equation} \tag{1}\label{1}A\times B=\pi_X^{-1}[A]\cap\pi_Y^{-1}[B] \end{equation} $$ holds, right?. Moreover, if $\mathfrak X$ and $\mathfrak Y$ are collections then we know that the euqlity $$ \begin{equation} \tag{2}\label{2}\left(\bigcap\mathfrak X\right)\cap\left(\bigcap\mathfrak Y\right)=\bigcap_{(X,Y)\in\mathfrak X\times\mathfrak Y}(X\cap Y) \end{equation} $$ holds. Finally if $f$ is a function from a set $H$ to a set $K$ we know that for any collection $\mathfrak L$ the equality $$ \begin{equation} \tag{3}\label{3}f^{-1}\left[\bigcap\mathfrak L\right]=\bigcap_{L\in\mathfrak L}f^{-1}[L] \end{equation} $$ holds.
So I asked to me if using eq. \eqref{1}, eq. \eqref{2} and eq. \eqref{3} is possibile to prove the equality $$ \begin{equation} \tag{4}\label{4}\left(\bigcap_{U\in\mathfrak U}U\right)\times\left(\bigcap_{V\in\mathfrak V}V\right)=\bigcap_{(U,V)\in\mathfrak U\times\mathfrak V}(U\times V) \end{equation} $$ where $\mathfrak U$ and $\mathfrak V$ are arbitrary collections. So to do this I put $$ M:=\bigcup\mathfrak U\quad\text{and}\quad N:=\bigcup\mathfrak V $$ and thus I observed the inclusion $$ \bigcap\mathfrak U\subseteq M $$ and the inclusion $$ \bigcap\mathfrak V\subseteq N $$ hold so that by \eqref{1}, \eqref{2} and \eqref{3} I argued the equality $$ \left(\bigcap_{U\in\mathfrak U}U\right)\times\left(\bigcap_{V\in\mathfrak V}V\right)=\pi_M^{-1}\left[\bigcap_{U\in\mathfrak U}U\right]\cap\pi_N^{-1}\left[\bigcap_{V\in\mathfrak V}V\right]=\left(\bigcap_{U\in\mathfrak U}\pi_M^{-1}[U]\right)\cap\left(\bigcap_{V\in\mathfrak V}\pi_N^{-1}[V]\right)=\bigcap_{(U,V)\in\mathfrak U\times\mathfrak V}\big(\pi_M^{-1}[U]\cap\pi_N^{-1}[V]\big)=\bigcap_{(U,V)\in\mathfrak U\times\mathfrak V}(U\times V) $$ since any element of $\mathfrak U$ is a subset of $M$ and since any element of $\mathfrak V$ is a subset of $N$.
Now let be $A$, $B$, $C$ and $D$ sets so that let's we put $$ P:=A\cup B\quad\text{and}\quad Q:=C\cup D $$ so that the inclusion $$ A,B\subseteq P $$ and the inclusion $$ C,D\subseteq Q $$ holds. So I observed that the equality holds $$ \begin{equation} \tag{5}\label{5} (A\cap B)\times(C\cap D)=\pi_P^{-1}[A\cap B]\cap\pi_Q^{-1}[C\cap D]=\big(\pi_P^{-1}[A]\cap\pi_P^{-1}[B]\big)\cap\big(\pi_Q^{-1}[C]\cap\pi_Q^{-1}[D]\big)=\big(\pi_P^{-1}[A]\cap\pi_Q^{-1}[C]\big)\cap(\pi_P^{-1}[B]\cap\pi_Q^{-1}[D]\big)=(A\times C)\cap(B\times D) \end{equation} $$
Now I am quite sure that \eqref{4} holds since it is differently proved in really many texts. However in a text (Teoria de Conjuntos by Fernando Hernandez) it is written that \eqref{5} is true surely if $A$ and $B$ are subset of any set $X$ and if $C$ and $D$ are subset of any set $Y$ but into this wiki article it is said apparently that \eqref{5} generally holds: so if $A$ and $B$ are not contained into a set $X$ and if $C$ and $D$ are not contained into a set $Y$ then this is not evident by more ordinary arguments (e.g. if $x$ is in $(A\cap B)\times(C\cap D)$ then...) to me so that I thought I take a mistake and thus I thought to put a question here where I ask if \eqref{5} generally holds and so if I well proved \eqref{4} and \eqref{5} since if \eqref{5} does not holds then it seems to me I do not well proved \eqref{4}. So could someone help me, please?
Your proofs of (5) and (4) are ok. (5) always hold. Though feasible (as you did when defining $P,Q$), it is useless to consider a set $X$ containing $A$ and $B$ and a set $Y$ containing $C$ and $D$ to prove (5) "by more ordinary arguments": $$\begin{align}(x,y)\in(A\cap B)\times(C\cap D)&\iff\left((x\in A)\land(x\in B)\right)\land\left((y\in C)\land(y\in D)\right)\\ &\iff\left((x\in A)\land(y\in C)\right)\land\left((x\in B)\land(y\in D)\right)\\ &\iff(x,y)\in(A\times C)\cap(B\cap D). \end{align}$$