Handling vacuous truths in proving elementary set theory So I was working on this simple problem:

For sets $A$ and $B$, prove that $A\subset B\Leftrightarrow A\cup B=B$.

Now I know how this proof works: First suppose $A\subset B$, then take $x\in A\cup B$, and so on. What I'm concerned about is how can we handle vacuous truths(or more specifically, empty sets). 
For example, let's show that if $A\cup B=B$, then $A\subset B$. We will most likely take $x\in A$. However, if $A=\varnothing$, then there is no such $x$. Another example is when we show the converse. Then, we will take $x\in A\cup B$. However, there is no such $x$ if $A\cup B$ is empty.
I know this may sound like nitpicking, but I'm not sure if I can just ignore those cases. I've seen numerous proofs here on MSE, but they seem to have no problem taking an element from a set that may be empty. Can anyone explain this to me? Thanks in advance.
 A: You are correct, and formally a lot of proofs should start with: If $A = \emptyset$, then it holds that $P$. So we assume that $A \neq \emptyset$ [..]
However in most situations the claim follows immediately for empty sets (because of vacuous truths), so it is common practice to omit the case $A = \emptyset$.
A: If $A=\varnothing$ then it is immediate that for every $x\in A$ we have $x\in B$.
This because no element of $A$ can be found for which this is false. 
This simply because $A$ has no elements.
You might get more grip "vacuous truth" if you look at the contrapositive statement.
The statement $x\notin B\implies x\notin\varnothing$ is evidently true, and is actually the "same" statement as $x\in\varnothing\implies x\in B$.
A: These proofs require a universal quantifier.
$$\forall x:p(x),$$
which can be read as $$p(x_1)\land p(x_2)\land\cdots p(x_n).$$
But in the vacuous case, the conjunction is simply
$$\text{true}.$$
Said differently, no element disproves the claim.

With the existential quantifier, we have a similar situation.
$$\exists x:p(x)$$
is
$$p(x_1)\lor p(x_2)\lor\cdots p(x_n),$$
which "degenerates" as
$$\text{false},$$
no element has the property.

You can compare to vacuous sums, which are defined as $0$, and vacuous products, $1$.
A: Those cases are handled implicitly.
When we say "Let $x \in A$" we are actually saying "For all $x$ such that $x \in A$" and that includes the vacuous case.  As all conditions are true for all $x$ that don't exist any statement about the vacuous state will be true.
"For all $x$ such that $x \in A$, it holds that $x\in A\cup B$".  That is true.  If there are such $x$ that follows by definition union. ANd if there are no such $x$ the statement is true as there are no such $x$ to falsify it.  
And for all $x$ it follows as all such $x \in A\cup B=B$ so for all such  $x: x\in B$. Again if there is no such $x\in A$ this vacuously follows as there are no $x$ to falsify it.
So you aren't ignoring anything.  You just arent explicitly stating these cases.
.....
Upshot:   "If $x$ does something A then $x$ does something B" means "for all $x$ such that $x$ does something A then $x$ does something B".  And that  will always be vacuously true if there are not $x$ that do something A.
