Prove or disprove elementary sets We have $ A \times B = (A \times B_1 ) \cup (A \times B_2)$
is it always true that:
Not sure (a) is sufficient proof and if (b) is adequate counter example
(a) $( A_1 \cap A_2 ) \times ( B_1 \cap B_2) = ( A_1 \times B_1) \cap ( A_2 \times B_2)$?
True; Proof:   
$ (A_1 \cap A_2 \times B_1) \cap (A_1 \cap A_2 \times B_2)$
$\Rightarrow (A_1 \times B_1 \cap  A_2 \times B_1) \cap (A_1 \times B_2\cap A_2 \times B_2)$
$\Rightarrow (A_1 \times B_1) \cap (A_2 \times B_2)$
(b) $(A_1 \cup A_2) \times (B_1 \cup B_2) = (A_1 \times B_1) \cup (A_2 \times B_2)$?
False; counterexample: 
Let $A_1 = B_1 = [0,1]$ and $A_2 = B_2 = [1,2]$
 A: The statement (a) is true. That is:
$$( A_1 \cap A_2 ) \times ( B_1 \cap B_2) = ( A_1 \times B_1) \cap ( A_2 \times B_2)$$
We can show this by using pointwise set-equality for sets $U, S$. That is,
$$U = S \leftrightarrow \forall x \;(x \in U \leftrightarrow x \in S)$$
In other words, assume an arbitrary $x$ is in $U$ and show it's in $S$ and vice versa, and you have set equality.
Applying this to the problem at hand, we have two cases. First assume there's an element of $( A_1 \cap A_2 ) \times ( B_1 \cap B_2)$. Then it's some pair $(a, b)$ with $a \in ( A_1 \cap A_2 )$ and $b \in ( B_1 \cap B_2)$. Now, because $a \in ( A_1 \cap A_2 )$, we also have $a \in A1$ and $a \in A_2$. Similarly for $b$. So then $(a, b) \in ( A_1 \times B_1)$ and $(a, b) \in ( A_1 \times B_1)$. So $(a, b) \in ( A_1 \times B_1) \cap ( A_2 \times B_2)$, concluding the case.
The next case is to assume some $(a, b) \in ( A_1 \times B_1) \cap ( A_2 \times B_2)$. Well, then, $(a, b) \in ( A_1 \times B_1)$ and $(a, b) \in ( A_1 \times B_1)$. But then $a \in A_1$ and $a \in A_2$. Similarly for $b$. So $a \in ( A_1 \cap A_2 )$, similarly for $b$. So then $(a, b) \in ( A_1 \cap A_2 ) \times ( B_1 \cap B_2)$, completing the proof.
The counter example in the question is actually not a counter example. The only element on both the left and right hand side is $(1, 1)$.
