Here is a lemma I deduced that describes a possible proof technique but I'm not sure what to prove with it.
If $F$ is a finite set of formulas, say $F$ distinguishes a finite set of theories $S$ if $\vert \{ T \cap F : T \in S \} \vert = \vert S \vert$.
Lemma: Suppose $S$ is a finite union-closed set of theories of odd cardinality and $A$ and $B$ are finite sets of formulas both distinguishing $S$. Then conditional on the union-closed sets conjecture, there exists $a \in A$, $b \in B$, and $T \in S$ such that $a \land b \in T$.
Corollary: if the members of $S$ are consistent extensions of $\text{ZFC}$, then there exists $a \in A$, $b \in B$ such that $\neg (a \land b)$ is not provable in $\text{ZFC}$.
The idea is that we have some consistent theories and by the union-closed sets conjecture more than half of them prove $a$ and more than half of them prove $b$, so at least one of them proves $a \land b$ but we don't know which. However, since they're all consistent extensions of $\text{ZFC}$, we know $\text{ZFC}$ can't prove $\neg(a \land b)$.
Is the above clear and correct?
Are there any undecidability results that can be proved with this method? Or anything at all interesting enough to justify calling it a "method" in the first place?
Your argument seems fine - but I don't think you need to use any open conjecture for this! If $T$ contains $a$ and $T'$ contains $b$, then just by the fact that the set of theories is union-closed, $T\cup T'$ contains $a$ and $b$, so it proves $a \land b$, so this sentence is consistent with ZFC. Right?
Unfortunately, I think the setup you have in mind is kind of trivial, since if you have a union-closed set of theories, then they are all consistent with ZFC if and only if the single theory which is the union of all of them is consistent with ZFC.