Velleman, exercise 23 section 3.3 on families of sets

I'm self-studying How to Prove It, by Daniel Velleman. The indicated exercise asks us to criticize an incorrect proof of the following incorrect theorem, and to provide a counterexample. I'd like to verify my answer. Thanks in advance for your help.

Incorrect Theorem: Suppose $\mathcal F$ and $\mathcal G$ are families of sets. If $\cup\mathcal F$ and $\cup\mathcal G$ are disjoint, then so are $\mathcal F$ and $\mathcal G$.

The incorrect proof goes like this (paraphrasing): Suppose $\cup\mathcal F$ and $\cup\mathcal G$ are disjoint and suppose $\mathcal F$ and $\mathcal G$ are not. We can choose a set $A$ which belongs to both $\mathcal F$ and $\mathcal G$. Now $A\in \mathcal F$ implies $A\subseteq\cup\mathcal F$ and $A\in \mathcal G$ implies $A\subseteq\cup\mathcal G$. But this contradicts the supposition that $\cup\mathcal F$ and $\cup\mathcal G$ are disjoint. So $\mathcal F$ and $\mathcal G$ must be disjoint. $\square$

The only flaw I can spot is on the sentence "But this contradicts..." because it tacitly supposes that $A\neq \emptyset$. My counterexample is $\mathcal F=\{\emptyset,\{1\}\}$ and $\mathcal G=\{\emptyset,\{2\}\}$, so that: $\mathcal F\cap \mathcal G=\{\emptyset\}$, $\cup\mathcal F=\{1\}$, $\cup\mathcal G=\{2\}$ and $\cup\mathcal F\cap\cup\mathcal G=\emptyset$.

Do you agree or I'm way off mark here?

Yes, I agree. You are entirely right. In fact, you spotted the only problem with that proof. If we add that $\emptyset\notin\mathcal{F}$ or $\emptyset\notin\mathcal{G}$, then the proof will be correct (and the statement will be true then, of course).