Finding flaw in proof

A good strategy is to to run through the proof with your counter-example. The issue is that while it is also true for $A = \emptyset$ that every element of $A$ is in both $\cup F$ and $\cup G$ this does not contradict the assertion that the sets are disjoint. Only if there is an element in $A$, there is a contradiction.

Note that this also shows the the only set that can be common to both $F$ and $G$ is the empty set.