Let $A,B,C$ be sets such that $\mathcal{P}(A) \cup \mathcal{P}(B) = \mathcal{P}(C)$. Then either $A=C$ or $B=C$.
Stuck with proof for $\forall A\forall B(\mathcal{P}(A)\cup\mathcal{P}(B)=\mathcal{P}(A\cup B)\rightarrow A\subseteq B \vee B\subseteq A)$