Prove $\bigcup (F\setminus G) \subseteq (\bigcup F) \setminus (\bigcup G)$ iff $\forall A \in (F\setminus G) \forall B\in G (A\cap B = \emptyset)$ This is an exercise from Velleman's "How To Prove It":

Prove that $\bigcup (F \setminus G) \subseteq (\bigcup F) \setminus (\bigcup G)$ iff $\forall A \in (F \setminus G) \forall B \in G (A \cap B = \emptyset)$.

I am uncertain about the use of variables in existential instantiation. If I say something like $\exists x P(x)$, is it okay to then continue using $x$ in the rest of the proof, or should I introduce a new variable $a$ such that $P(a)$? Also, when using contradiction, is it necessary to indicate that I am doing so? As I am self-studying, I would greatly appreciate other comments as well. Thanks in advance!
Proof: Suppose $\bigcup (F \setminus G) \subseteq (\bigcup F) \setminus (\bigcup G)$. Let $A \in (F \setminus G) $ and $B \in G$ be arbitrary. Now suppose $\exists x (x \in A \cap B)$. Since $x \in A$ and $A \in (F \setminus G)$, it follows by definition that $x \in \bigcup (F \setminus G)$. Since  $x \in \bigcup (F \setminus G)$ and $\bigcup (F \setminus G) \subseteq (\bigcup F) \setminus (\bigcup G)$, $x \in (\bigcup F) \setminus (\bigcup G)$. Thus, $x \in (\bigcup F) $ and $x \notin (\bigcup G) $. But since $x \in B$ and $B \in G$, 
 $x \in \bigcup G$. Thus, we have $x \in \bigcup G$ and $x \notin \bigcup G$, which is a contradiction. So $\forall x (x \notin A \cap B)$ and $A \cap B = \emptyset$. Since $A$ and $B$ were arbitrary, it follows that $\forall A \in (F \setminus G) \forall B \in G (A \cap B = \emptyset)$.
Suppose $\forall A \in (F \setminus G) \forall B \in G (A \cap B = \emptyset)$. Let $x \in \bigcup (F \setminus G)$ be arbitrary. Since $x \in \bigcup (F \setminus G)$, we can choose some $W \in (F \setminus G)$ such that $x \in W$. Since $x \in W$ and $W \in F$, it follows that $x \in \bigcup F$ by definition. Suppose $x \in \bigcup G$. We can then choose a $V \in G$ such that $x \in V$. But then we have $W \in (F \setminus G)$, $V\in G$, and $x \in W \cap V$. This is a contradiction because it was given that $\forall A \in (F \setminus G) \forall B \in G (A \cap B = \emptyset)$. Thus, $x \notin \bigcup G$. Therefore, if $x \in \bigcup (F \setminus G)$, then $x \in (\bigcup F) \setminus (\bigcup G)$. Since $x$ was arbitrary, $\bigcup (F \setminus G) \subseteq (\bigcup F) \setminus (\bigcup G)$. $\square$
 A: Great job on your proof! 
It's not strictly necessary to note at the beginning that you'll proceed by contradiction. Some people use "we'll proceed by contradiction" before the argument but just by seeing "suppose that there is an $x$ such that $x \in A \cap B$" most people will already know what you're doing.
About variables in existential instantiation, yes, you can just keep using them because it makes sense. Strictly speaking, it formally doesn't make sense to keep talking about an $x$ that was previously instantiated between those parentheses, but remember mathematical proofs are not computer programs but something a human will read. I prefer to write something like "Assume we can take an $x$ such that $P(x)$" since natural language lets you forget about those low-level logic issues. 
With respect to style, there's some things that could be omitted in your proof to make it shorter. Of course, this can get into dangerous territory, so I suggest reading a lot of proofs from different areas, like number theory, to get a sense of how people write. Eventually you'll develop your own style.
