Proving set identities: empty set case. I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$A\cap(B \cup C) = (A \cap B) \cup (A \cap C).$$
I understand that this can be done by proving both $$A\cap(B \cup C) \subset (A \cap B) \cup (A \cap C) \ \text{and} \ (A \cap B) \cup (A \cap C) \subset A\cap(B \cup C)  $$ hold true. 
We can do this by letting $x$ be an arbitrary element of $A\cap(B \cup C)$ and showing that it is an element of $(A \cap B) \cup (A \cap C)$, and vice versa. 
But what about when $A \cap (B \cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A \cap (B \cup C)$ since there are none. However, I am aware that $A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$ is trivially true in this case.
In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not? 
If so, some suggestions as to how to incorporate them into proofs would be helpful :-).
 A: If $D$ is the empty set then the following statement is always true:$$\text{If }x\in D\text{ then }\cdots$$no matter what the dots are standing for.
It is false that $x\in D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.
So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.
A: Since the empty set is a subset of any set, there is no need of including that in a formal proof. 
However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets . 
A: In order to show that for two sets $X$ and $Y$ it holds that $X\subseteq Y$, you have to prove that

for every $x$, if $x\in X$, then $x\in Y$.

Note that a statement of the form “if $\mathscr{A}$ then $\mathscr{B}$” is true when

either $\mathscr{A}$ is false or both $\mathscr{A}$ and $\mathscr{B}$ are true

If $X$ is the empty set, then “$x\in X$” is false for every $x$; hence “if $x\in X$ then $x\in Y$” is true.
The phrase “take an arbitrary element $x\in X$” is possibly misleading, but its intended meaning is “suppose $x\in X$”.
