Check if $\mathcal A $ and$ \mathcal B $ : $\bigcup (\mathcal A \cap \mathcal B) = \bigcup \mathcal A \cap \bigcup \mathcal B$ 
Check if for any two set families $\mathcal A $ and $\mathcal B $ the
  following is true:  $\bigcup (\mathcal A \cap \mathcal B) = \bigcup
 \mathcal A \cap \bigcup \mathcal B$

First of all I considered an example: $\mathcal A = \{ \{1,2\}, \{1,3\} \}$ and $\mathcal B = \{\{1,2\},\{3,5\}\}$
Now, $\mathcal A \cap \mathcal B = \{1,2 \}$, and so $\bigcup(\mathcal A \cap \mathcal B) =\{1,2 \}$ Whereas, on the other hand, $\bigcup \mathcal A= \{1,2,3 \} $ and $\bigcup \mathcal B = \{1,2,3,5 \}$ and so their intersection is $\{1,2,3 \}$ and so my thesis is - the statement is not true. But now, when I start to evaluate it using the Axiom of Extensionality, I get: $$(\exists X)((X\in \mathcal A \land X\in \mathcal B )\land x \in X)$$
$$\iff(\exists X)(X \in \mathcal A\land x\in X) \land (\exists X)(X \in \mathcal B \land x\in X)$$
$$\iff x \in \bigcup \mathcal A \land x \in \bigcup \mathcal B \iff x \in (\bigcup \mathcal A \cap \bigcup \mathcal B)$$ 
So, on the one hand the example I provided shows that the theorem is false, but on the other hand, the definitions says that it is actually true. Therefore I must have erred somewhere in my reasoning but I can't see where. Could you steer me towards this error?
 A: The counterexample you provided is solid and disproves the theorem right away, so there must be an error in your equivalence chain proving the theorem. The error is in the statement
\begin{align}
&(\exists X)((X\in \mathcal A \land X\in \mathcal B )\land x \in X)\\
\iff&(\exists X)(X \in \mathcal A\land x\in X) \land (\exists X)(X \in \mathcal B \land x\in X).
\end{align}
Those expressions are not equivalent. The first expression states that there exists a set $X$ which is in both $\mathcal A$ and $\mathcal B$ such that also $x\in X$. Whereas the second expression says that there exists a set $X$ from just $\mathcal A$ which contains $x$, but also a (possibly different) set $X$ from just $\mathcal B$ that contains $x$.
So the point is that in the second expression you don't relate families $\mathcal A$ and $\mathcal B$ in any way. A way to perhaps make this clearer is to realise that the statement $$(\exists X)(X \in \mathcal A\land x\in X) \land (\exists X)(X \in \mathcal B \land x\in X)$$
is equivalent to
$$(\exists X)(X \in \mathcal A\land x\in X) \land (\exists Y)(Y \in \mathcal B \land x\in Y),$$
which can' be equivalent to 
$$(\exists X)((X\in \mathcal A \land X\in \mathcal B )\land x \in X).$$
A: Generally, when you have to prove or disprove a statement, a single counterexample (along with the proof that it is indeed a counterexample) works. Your counterexample indeed works, and you can even make it simpler with $\mathcal A=\{\{1\},\{2\}\}$ and $\mathcal B=\{\{1,2\}\}$.
Now, the general case you have a mistake that $\exists$ does not distribute over $\land$. It does distribute over $\lor$, though. In other words, $$\exists x(\varphi(x)\land\psi(x)\not\equiv(\exists x\varphi(x))\land(\exists x\psi(x)).$$
This is exactly the difference between "The intersection of these two sets is not empty" and "These two sets are not empty".
A: $$\begin{aligned}x\in\bigcup\left(\mathcal{A}\cap\mathcal{B}\right) & \iff\exists y\left[y\in\mathcal{A}\cap\mathcal{B}\wedge x\in y\right]\\
 & \iff\exists y\left[y\in\mathcal{A}\wedge y\in\mathcal{B}\wedge x\in y\right]\\
 & \overset{\text{not the opposite}}{\implies}\exists y\left[y\in\mathcal{A}\wedge x\in y\right]\wedge\exists y\left[y\in\mathcal{B}\wedge x\in y\right]
\end{aligned}
$$
The last expression is equivalent with $x\in\left(\bigcup\mathcal{A}\right)\cap\left(\bigcup\mathcal{B}\right)$.
