When letting an element be an arbitrary member of a set $A$ do I have to account for $A =\emptyset$? When  letting an element be an  arbitrary member of a set $A$ do I always have to account for $A =\emptyset$?
E.g., if I wanted to prove (an incorrect) theorem 
$$ \forall A,B,C,D : A \times B \subset C \times D \implies (A \subset C \wedge B \subset D).$$
I would let $x$ be an arbitrary element of $A$ but I cannot proceed with the proof without considering the case when A is $\emptyset$. Do I always have to do this whenever I pick an arbitrary element of a set?
 A: You have to account for it. However, this is often automatically done, because any statement in the form $$[\forall x\in A, P(X)]\equiv [\forall x,(P(x)\vee\neg x\in A)]$$ is always true when $A=\emptyset$.
You must be careful when you make an existence claim, because any satement in the form $$[\exists x\in A,\ P(x)]\equiv [\exists x,\ (x\in A\wedge P(x))]$$ is always false when $A=\emptyset$.
A: Generally, yes. Usually, the theorem is easy to prove when the set is empty, so just start off by handling the case where $A$ is empty, then assume $A$ is nonempty so you can pick $x\in A$. 
EDIT- Though there are some cases where you don't have to handle the two cases. For instance, the definition of $A\subset B$ is $x\in A \implies x\in B$; to prove $A \subset B$, you simply show that if we have some $x\in A$, then it follows that $x\in B$ (regardless of whether $A$ happens to be empty). Generally, just ask yourself: if you don't handle the possibility that your set is empty, is your proof incomplete? Have you failed to prove your result in the case that $A=\emptyset$?
Your example statement would fall into the second category. 
A: Yes, you do have to do that everytime.
It could even be a method to prove that a given set $E$ is empty :
Let $x$ be in $E$... and then you get to a contradiction. And you conclude that $E=\emptyset$.
