Prove that if $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$ then either $A \subseteq B$ or $B \subseteq A$. Prove that for any sets $A$ or $B$, if $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$ then either $A \subseteq B$ or $B \subseteq A$. ($\mathcal P$ is the power set.)
I'm having trouble making any progress with this proof at all. I've assumed that $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$, and am trying to figure out some cases I can use to help me prove that either $A \subseteq B$ or $B \subseteq A$. 
The statement that $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$ seems to be somewhat useless though. I can't seem to make any inferences with it that yield any new information about any of the sets it applies to or elements therein. The only "progress" I seem to be able to make is that I can conclude that $A \subseteq A \cup B$, or that $B \subseteq A \cup B$, but I don't think this gives me anything I don't already know. I've tried going down the contradiction path as well but I haven't been able to find anything there either.
I feel like I am missing something obvious here though...

Hint: Try instead to prove the contrapositive: 

If $A \nsubseteq B$ and $B \nsubseteq A$, then $\mathcal{P} ( A ) \cup \mathcal{P} ( B ) \neq \mathcal{P} ( A \cup B )$.

Remember that $E \nsubseteq F$ means that there is an element of $E$ which is not an element of $F$.
