Cross product intersection sets Here's what I'm, trying to prove.
Let $A, B, C$ be non-empty sets. Prove that $A \times (B \cap C)  \subseteq  (A \times B) \cap (A \times C)$
First I would need to prove $A \times (B \cap C)  = (A \times B) \cap (A \times C)$
If I used distributivity rule then I get: $(A \times B) \cap (A \times C)  = (A \times B) \cap (A \times C)$
I feel like using distributivity here is wrong because both sides equal each other, which becomes a proper subset. Am I wrong? 
Then I would get $(A \times B) \cap (A \times C)  \subset (A \times B) \cap (A \times C)$
 A: Edit: From the comment, it seems you are being asked to show equality (distributivity of $$A \times (B \cap C)  =  (A \times B) \cap (A \times C)$$
and you can do this by "element chasing" I.e., show that both the following hold: $$(x,y) \in \Big[A \times (B \cap C)\Big]  \implies (x, y) \in\Big[ (A \times B) \cap (A \times C)\Big] $$ $$ (x,y) \in\Big[ (A \times B) \cap (A \times C)\Big] \implies (x, y) \in \Big[A \times (B \cap C)\Big]$$and you're done.  But note that you cannot use what you are asked to prove (distributivity of the cross product). Use the definitions of the cross-product, and the definition of set intersection to prove the above (and also distributivity over conjunction/set intersection).
You can also start with unpacking the definition of $A \times (B\cap C)$ using set-builder notation, and through step by step equivalency, arrive at the set defining $(A \times B) \cap (A \times C)$, showing that we do in fact have that equality holds.
For example:
$$\begin{align} A \times (B\cap C) & = \{(x, y)\mid x \in A \land y \in (B \cap C) \}\\ \\  
& = \{(x, y)\mid x \in A \land (y \in B \land y \in C)\} \\ \\  
& = \qquad \vdots \\ \\ 
& = \{(x, y)\mid (x \in A \land y \in B) \land (x \in A \land y \in C)\} \\ \\
&= (A \times B) \cap (A \times C)\end{align}$$
