Proving that $\bigcup\mathcal{A}\times\bigcap\mathcal{B}\subseteq\bigcup\{a\times b\mid a\in\mathcal{A}\land b\in\mathcal{B}\}$ is always true The problem is to prove that the following expression is true for any families of sets $\mathcal{A}$ and $\mathcal{B}$ that are not empty.
$$\bigcup\mathcal{A}\times\bigcap\mathcal{B}\subseteq\bigcup\{a\times b\mid a\in\mathcal{A}\land b\in\mathcal{B}\}$$
Our definition of cartesian product is that $A \times B = \{  (a,b) | a\in A \land b \in B \}$ and and ordered pair $(a,b)=(c,d)$ if and only if $a=c$ and $b=d$. So we haven't seen Kuratowski's definition of ordered pairs yet.
I'm not searching for a complete solution, the problem is I don't know where to start. I know what a family of sets is but I can't see what is a sum of all cartesian products $a \times b$ (right side of the equation). How should I start approaching such a problem? I'm familiar with easier proofs with sets but these cartesian products and operations  with $\bigcup$ and $\bigcap$ are new to me.
 A: Here is a proof where we just expand the definitions, and then use the laws of logic to simplify and see where that leads us.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Tag}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$Looking at both sides of $\;\bigcup\mathcal{A}\times\bigcap\mathcal{B}\subseteq\bigcup\{a\times b\mid a\in\mathcal{A}\land b\in\mathcal{B}\}\;$, it is clear that each side is a set of ordered pairs.  Therefore we calculate which pairs $\;(x,y)\;$ are in the most complex side, the right hand side, and simplify:
$$\calc
(x,y) \in \bigcup\{a\times b\mid a\in\mathcal{A}\land b\in\mathcal{B}\}
\calcop\equiv{definition of $\;\bigcup\;$}
\langle \exists P : (x,y) \in P : P \in \{a\times b\mid a\in\mathcal{A}\land b\in\mathcal{B}\} \rangle
\calcop\equiv{definition of set binder notation}
\langle \exists P : (x,y) \in P : \langle \exists a,b : a\in\mathcal{A}\land b\in\mathcal{B} : P = a\times b \rangle \rangle
\calcop\equiv{logic: exchange $\;\exists\;$ quantifications -- to exploit $\;P = \ldots\;$}
\langle \exists a,b : a\in\mathcal{A}\land b\in\mathcal{B} : \langle \exists P : (x,y) \in P : P = a\times b \rangle \rangle
\calcop\equiv{logic: one-point rule, i.e., substitute for $\;P\;$}
\langle \exists a,b : a\in\mathcal{A}\land b\in\mathcal{B} : (x,y) \in a\times b \rangle
\calcop\equiv{definition of $\;\times\;$}
\langle \exists a,b : a\in\mathcal{A}\land b\in\mathcal{B} : x \in a \land y \in b\rangle
\calcop\equiv{logic: split into independent $\;\exists\;$ quantifications}
\langle \exists a : a\in\mathcal{A} : x \in a \rangle \;\land\; \langle \exists b : b\in\mathcal{B} : y \in b \rangle
\calcop\equiv{definition of $\;\bigcup\;$, twice}
x \in \bigcup \mathcal A \;\land\; u \in \bigcup \mathcal B
\calcop\equiv{definition of $\;\times\;$}
(x,y) \in \bigcup \mathcal A \times \bigcup \mathcal B
\endcalc$$
So by set extensionality, this proves
$$
\bigcup\{a\times b\mid a\in\mathcal{A}\land b\in\mathcal{B}\} \;=\; \bigcup \mathcal A \times \bigcup \mathcal B
$$
(as you correctly observed in a comment to the first answer).
So all that is left to prove is
$$
\bigcup \mathcal A \times \bigcap \mathcal B \;\subseteq\; \bigcup \mathcal A \times \bigcup \mathcal B
$$
This follows directly from two other theorems, one of which is $\;\bigcap S \;\subseteq\; \bigcup S\;$, which -- if treated like the proof above -- turns out to come down to the $\;\forall \Rightarrow \exists\;$ law of logic.
A: Consider $(a,b)\in \bigcup\mathcal{A} \times \bigcap\mathcal{B}$.
We have $a\in \bigcup\mathcal{A}$ so there exists $A\in\mathcal{A}$ such that $a\in A$. Let's note it $A_0$.
We have $b\in\bigcap\mathcal{B}$ so for all $B\in\mathcal{B}$ we have $b\in B$.
But then, $(a,b)\in A_0 \times B$ for any $B\in\mathcal{B}$, so
$$(a,b) \in \bigcup_{A\in\mathcal{A}, B\in\mathcal{B}}A\times B.$$
Actually, you have
$$\bigcup_{A\in\mathcal{A}, B\in\mathcal{B}}A\times B = \bigcup\mathcal{A} \times \bigcup\mathcal{B} \supset \bigcup\mathcal{A}\times\bigcap\mathcal{B},$$
with equality if and only if all $B$'s in $\mathcal{B}$ are the same.
