Prove $\bigcup\mathcal{F}$ and $\bigcup\mathcal{G}$ are disjoint iff for all $A \in \mathcal{F}$ and $B \in \mathcal{G}$, A and B are disjoint. Prove that $\bigcup\mathcal{F}$ and $\bigcup\mathcal{G}$ are disjoint iff for all $A \in\mathcal{F}$ and $B \in \mathcal{G}$, $A$ and $B$ are disjoint.
Scratch work: Question asks to prove iff so I will start out trying to prove ($\rightarrow$). 
To make clear how I should approach the proof, I will analyze the logical form of the goal for ($\rightarrow$). Thus, I get $$ \tag1 \lnot \exists  x (x\in \cup \mathcal{F} \land x \in \cup \mathcal{G}) \rightarrow \forall A \in \mathcal{F} \forall B \in \mathcal{G} (A \cap B = \emptyset)$$ This forms suggests I should begin by assuming the antecedent. The antecedent can be translated to the following $$ \tag2 \forall x (x \in \cup \mathcal{F} \rightarrow x \notin \cup \mathcal{G}) $$ Thus I must prove  $$ \tag3 \forall A \in \mathcal{F} \forall B \in \mathcal{G}(A \cap B = \emptyset)$$ Again, I may rewrite the A $\cap$ B = $\emptyset$ as $$ \tag4 \forall A \in \mathcal{F} \forall B \in \mathcal{G} \forall x (x \in A \rightarrow x \notin B)$$ Thus, I may assume an arbitrary $A \in \mathcal{F}$, an arbitrary $B \in \mathcal{G}$ and an arbitrary $x \in A$. Now I will show that $x \notin B$. 
Because $x \in A$ and $A \in \mathcal{F}$, $x \in \bigcup\mathcal{F}$. By (2), $x \notin \bigcup\mathcal{G}$. Because $B \in \mathcal{G}$ and $x \notin \bigcup\mathcal{G}$, $x \notin B$. 
Proof. ($\rightarrow$) Suppose $\bigcup \mathcal{F}$ and $\bigcup \mathcal{G}$ are disjoint. Suppose $A \in \mathcal{F}$, $B \in\mathcal{G}$, and $x \in A$. Since x $\in$ A and A $\in$ $\mathcal{F}$, x $\in$ $\cup$ $\mathcal{F}$. Because $\bigcup \mathcal{F}$ and $\bigcup \mathcal{G}$ are disjoint, $x \notin \bigcup \mathcal{G}$. Because $B \in \mathcal{G}$, $x \notin B$. Because $x$ was an arbitrary element of $A$, $A$ and $B$ are disjoint. Since we assumed arbitrary sets $A$ and $B$, this is true for all sets $A$ and $B$. Thus, if $\bigcup\mathcal{F}$ and $\bigcup\mathcal{G}$ are disjoint, all sets $A$ and $B$ are disjoint. $\square$
Does my proof say all that it needs to say or does it say too much? I have heard that it is sometimes not necessary to point out that some sets or elements were arbitrary as it can be assumed. Also, I left out the proof for ($\leftarrow$) as I'm only concerned with this portion right now. 
 A: The proof is a little wordy and a little harder to follow than necessary. At a beginning level I’d present the same argument more like this:

Suppose that $\bigcup\mathscr{F}$ and $\bigcup\mathscr{G}$ are disjoint. Let $A\in\mathscr{F}$ and $B\in\mathscr{G}$; we want to show that $A\cap B=\varnothing$. Suppose that $x\in A$; $A\in\mathscr{F}$, so $x\in A\subseteq\bigcup\mathscr{F}$, and therefore $x\in\bigcup\mathscr{F}$. And $\mathscr{F}$ is disjoint from $\bigcup\mathscr{G}$, so $x\notin\bigcup\mathscr{G}$. Finally, $B\in\mathscr{G}$, so $\bigcup\mathscr{G}\supseteq B$, and therefore $x\notin B$. Thus, $A\cap B=\varnothing$, as desired. $\dashv$

If I were starting from scratch, however, I would actually prove the contrapositive: if there are $A\in\mathscr{F}$ and $B\in\mathscr{G}$ such that $A\cap B\ne\varnothing$, then $\bigcup\mathscr{F}$ and $\bigcup\mathscr{G}$ are not disjoint.

Suppose that $A\in\mathscr{F}$, $B\in\mathscr{G}$, and $A\cap B\ne\varnothing$; then there is an $x\in A\cap B$. But then $x\in A\subseteq\bigcup\mathscr{F}$, since $A\in\mathscr{F}$, and similarly $x\in B\subseteq\bigcup\mathscr{G}$, so $\bigcup\mathscr{F}$ and $\bigcup\mathscr{G}$ are not disjoint.

I would also prove the contrapositive for the other direction.
