Critique my proof on correctness, structure, etc.
Proof. Suppose $p = (x,y)$ is an arbitrary element.
($\subseteq$)
Let $p \in A \times (B \cap C)$. By definition of cartesian product, $x \in A$ and $y \in B \cap C$. Thus, $y \in B$, $y \in C$ and $p \in (A \times B)$, $p \in (A \times C)$. By definition of $(A \times B) \cap (A \times C)$, $\ p \in (A \times B) \cap (A \times C)$ and because $p$ is arbitrary, we can say that $A \times (B \cap C) \subseteq (A \times B) \cap (A \times C)$.
($\supseteq$)
Let $p \in (A \times B) \cap (A \times C)$. Thus, $p \in (A \times B)$ and $p \in (A \times C)$. By definition of cartestian product, $x \in A$ and $y \in B$, $y \in C$. By definition of $A \times (B \cap C)$, $\ p \in A \times (B \cap C)$ and because $p$ is arbitrary, we can say that $(A \times B) \cap (A \times C) \subseteq A \times (B \cap C)$.
$\therefore$ $A \times (B \cap C) \subseteq (A \times B) \cap (A \times C)$ and $(A \times B) \cap (A \times C) \subseteq A \times (B \cap C)$, so $A \times (B \cap C) = (A \times B) \cap (A \times C)$.
Best Answer
I prefer to solve this kind of exercise by making use of the definitions and properties involved: \begin{align*} A\times(B\cap C) & = \{(x,y) \mid (x\in A)\wedge(y\in B\cap C)\}\\\\ & = \{(x,y) \mid (x\in A)\wedge(y\in B)\wedge(y\in C)\}\\\\ & = \{(x,y) \mid (x\in A)\wedge(x\in A)\wedge (y\in B)\wedge(y\in C)\}\\\\ & = \{(x,y) \mid ((x\in A)\wedge (y\in B))\wedge((x\in A)\wedge(y\in C))\}\\\\ & = \{(x,y) \mid (x\in A)\wedge(y\in B)\}\cap\{(x,y) \mid (x\in A)\wedge(y\in C)\}\\\\ & = (A\times B)\cap(A\times C) \end{align*}
Hopefully this helps !