Suppose $F$ and $G$ are families of sets. Prove that $(\bigcup F) \setminus (\bigcup G) \subseteq \bigcup (F \setminus G)$. This is Velleman's exercise 3.4.20.a:
Suppose $F$ and $G$ are families of sets. Prove that $(\bigcup F) \setminus (\bigcup G) \subseteq \bigcup (F \setminus G)$.
This is my proof but I'm suspicious of it being correct:
Proof. Suppose $x$ is an arbitrary element of $(\bigcup F) \setminus (\bigcup G)$. Therefore we can choose some $A \in F$ such that $x \in A$. And for all A ∈ G, x ∉ A. Now suppose ∀A ∈ F\G(x ∉ A). Thus whether A ∈ G or A ∉ G we get x ∉ A which is a contradiction. Therefore ∃A ∈ F\G(x ∈ A). Since x was arbitrary, $(\bigcup F) \setminus (\bigcup G) \subseteq \bigcup (F \setminus G)$.
Please tell me if my proof is wrong plus how to fix it (preferably write a proof in Velleman's expository style). 
Thanks in advance.
 A: You fixed $A$ earlier in the argument, but then proceeded to use the symbol $A$ to stand for an arbitrary element in $G$. You should use a different letter. Also, I don't understand the line "Thus whether $A\in G$ or $A\not\in G$ we get $x\not\in A$ which is a contradiction." Are you saying that $x\not\in A$ for all $A\not\in G$?
Here's how I would go about the proof:
Suppose $x\in(\cup F)\setminus(\cup G)$. Then $x\in\cup F$, so there exists $A\in F$ satisfying $x\in A$. Since $x\not\in\cup G$ and $x\in A$, we know that $A\not\in G$. Therefore $A\in F\setminus G$, which implies that $x\in A \subseteq \cup (F\setminus G)$. $\square$
I know you're following the notation of your textbook, but I would much prefer to use $\mathscr{F}$ and $\mathscr{G}$ instead of $F$ and $G$. The psychological difference between using script letters for sets of sets, capital letters for sets of elements, and lower case letters for elements is helpful. (Even though, technically, everything is a set.) Then it would read:

Given families of sets $\mathscr{F}$ and $\mathscr{G}$, we have $(\cup\mathscr{F})\setminus(\cup\mathscr{G}) \subseteq \cup(\mathscr{F}\setminus\mathscr{G})$.

Proof:
Suppose $x\in(\cup \mathscr{F})\setminus(\cup \mathscr{G})$. Then $x\in\cup\mathscr{F}$, so there exists $F\in\mathscr{F}$ satisfying $x\in F$. Since $x\not\in\cup\mathscr{G}$ and $x\in F$, we know that $F\not\in\mathscr{G}$. Therefore $F\in \mathscr{F}\setminus\mathscr{G}$, which implies that $x\in F \subseteq \cup (\mathscr{F}\setminus\mathscr{G})$. $\square$
