Proving that if $A\times (B \cup C)=A\times (B \cap C)$ then $A=\emptyset$ or $B=C$ 
Prove that if $A\times (B \cup C)=A\times (B \cap C)$ then $A=\emptyset$ or $B=C$.

Proof by contradiction: 
Givens: $A\times (B \cup C)=A\times (B \cap C)$, $A\neq\emptyset$, $B\neq C$.
Suppose $x\in A, y \in B$ so $y\not\in C$, then $(x,y)\in A\times (B \cup C)$.
So $(x,y)\in A\times (B \cap C)$ so $y\in B\cap C$ but since $B\neq C$ then $y\in B\cap C=\emptyset$ is a contradiction, thus: 
If $A\times (B \cup C)=A\times (B \cap C)$ then $A=\emptyset$ or $B=C$.
Is anything missing?
 A: Almost correct, but without contradiction it's easier.
Suppose $A\ne\emptyset$ and let $x\in A$. Consider $y\in B$: then $(x,y)\in A\times(B\cup C)$, so $(x,y)\in A\times(B\cap C)$. Therefore $y\in B\cap C$. This means $B\subseteq B\cap C$ and so $B\subseteq C$. By symmetry, $C\subseteq B$.
What's wrong in your argument? If $y\in B$, $y\notin C$ is not certain. However, you can take $y\in B\setminus C$ or $y\in C\setminus B$ and such an element certainly exists because $B\ne C$.
A: This looks mostly right, but you have to be a bit more careful. If you assume $B \neq C$, then you don't necessarily know $B\cap C = \emptyset$. Further, you don't know that there is necessarily $y \in B$ such that $y \not\in C$ (it could be the case that $B \subsetneq C$). It is true that either


*

*$\exists y \in B$ such that $y \not\in C$ 


OR


*

*$\exists y \in C$ such that $y \not\in B$


which is enough to make your proof work.
A: Your argument is wrong. It is not true that $B \cap C = \emptyset$.
$B \neq C$ is equivalent on saying that $( B \setminus C \neq \emptyset$ or $B \setminus C \neq \emptyset)$.
In the first case you can pick $y \in B$ and $y \notin C$, and use your argument.
In the second case you can use the same argument exchanging the roles of $B,C$.
A: Assumed is $A\neq\emptyset$ and $B\neq C$. 
The first assumption guarantees the existence of some $x\in A$. 
The second guarantees the existence of some $y$ that belongs to $B\cup C$
and does not belong to $B\cap C$ (there is an element in $B$ that is not in $C$ or there is an element in $C$ that is not in $B$).
Then $\langle x,y\rangle\in A\times\left(B\cup C\right)$ and $\langle x,y\rangle\notin A\times\left(B\cap C\right)$
so $A\times (B\cup C) \neq A\times\left(B\cap C\right)$
A: This is not feedback on your proof, but this is just to show an alternative style of proof: expand the definitions, and then use the laws of logic to simplify the result, and see where that leads you.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$We calculate as follows:
$$\calc
\tag{*} A \times (B \cup C) = A \times (B \cap C)
\calcop={set extensionality; definitions of $\;\times\;$, twice, and of $\;\cup\;$ and $\;\cap\;$}
\langle \forall x,y :: x \in A \land (y \in B \lor y \in C) \;\equiv\; x \in A \land (y \in B \land y \in C) \rangle
\calcop={logic: extract common conjunct $\;x \in A\;$ from both sides of $\;\equiv\;$}
\langle \forall x,y :: x \in A \;\Rightarrow\; (y \in B \lor y \in C \;\equiv\; y \in B \land y \in C) \rangle
\calcop={logic: write $\;P \Rightarrow Q\;$ as $\;\lnot P \lor Q\;$; split into independent quantifications}
\langle \forall x :: x \not\in A \rangle \;\lor\; \langle \forall y :: y \in B \lor y \in C \;\equiv\; y \in B \land y \in C \rangle
\calcop={logic: apply the 'golden rule' $\;P \lor Q \;\equiv\; P \land Q \;\equiv\; P \;\equiv\; Q\;$}
\langle \forall x :: x \not\in A \rangle \;\lor\; \langle \forall y :: y \in B \;\equiv\; y \in C \rangle
\calcop={definition of $\;\emptyset\;$; set extensionality}
\tag{**} A = \emptyset \;\lor\; B = C
\endcalc$$
Therefore we've not only proven that $\ref{*}$ implies $\ref{**}$, but we've proven them equivalent.
Yes, it's more symbols than the other answers-- but the good news is that all steps are of the type "there's really only one thing you can do", so if you know the laws of logic, it is almost trivial to write down.
