Prove: $A × (B ∩ C) = (A × B) ∩ (A × C).$ What I know:

-I need to prove the sets are equal, that is, prove the LH side is a subset of the RH side, and vice versa.
-$A × (B ∩ C)$ is the set of all ordered pairs with first entry an element of $A$, and second entry an element of $(B ∩ C)$.

-$(A × B)$ is the set of all ordered pairs with first entry an element of $A$, and second entry an element of $B$.
-$(A × C)$ is the set of all ordered pairs with first entry an element of $A$, and second entry an element of $B$.
So $(A × B) ∩ (A × C)$ is the set of ordered pairs that are both in $(A × B)$ and $(A × C)$.

How can I link these concepts together? I don't need a complete solution; getting started is the hardest part!
You need to "element chase"; that is, suppose $(x,y)\in A\times(B\cap C)$ and then show you must have $(x,y)\in (A\times B)\cap(A\times C)$, then do the other direction as well. I'll do one direction for you and you can try the other yourself.

Suppose $(x,y)\in A\times(B\cap C)$. Then $x\in A$ and $y\in B\cap C$, meaning $y\in B$ and $y\in C$. Because $y\in B$, $(x,y)\in A\times B$, and because $y\in C$, $(x,y)\in A\times C$. Therefore $(x,y)\in(A\times B)\cap(A\times C)$. This tells us $A\times(B\cap C)\subseteq (A\times B)\cap(A\times C)$.
Proof:
$$\begin{align}
A\times(B\cup C)&amp;=\{(x,y):x\in A,y\in B\cup C\}\\
&amp;=\{(x,y):x\in A,y\in B\text{ or }x\in A,y\in C\}\\
&amp;=\{(x,y):(x,y)\in A\times B\text{ or }(x,y)\in A\times C\}\\  
&amp;=(A\times B)\cup(A\times C)     
\end{align}$$
Therefore,  $A\times (B\cup C) = (A\times B)\cup(A\times C)$.