Set Theory proof from "How to Prove It" I am having difficulties with this proof--Hints will be appreciated.
Prove that for any sets $A$ and $B$, if $\mathcal{P}(A) \cup \mathcal{P}(B) =  \mathcal{P}(A\cup B)$ then either $A$ is a subset of $B$ or $B$ is a subset of $A$.
 A: Hint: We have $A\cup B\in\mathcal P(A\cup B)$. By hypothesis $\mathcal P(A)\cup\mathcal P(B)=\mathcal P(A\cup B)$, we obtain $A\cup B\in\mathcal P(A)\cup\mathcal P(B)$. Now, by definition of union, we have $A\cup B\in\mathcal P(A)$ or $A\cup B\in\mathcal P(B)$, and so $A\cup B\subseteq A$ or $A\cup B\subseteq B$ by definition of power set. In particular, we know that $A\subseteq A\cup B$ and $B\subseteq A\cup B$; thus the claim follows.
A: Suppose that we have the assumption but neither of the inclusions holds. Then $A \nsubseteq B$ so there is some $a \in A, a \notin B$ and also $B \nsubseteq A$ and so there is some $b \in B, b \notin A$. Now consider $\{a,b\} \subseteq A \cup B$, so $\{a,b\} \in \mathcal{P}(A \cup B)$. 
A: As $A \cup B \in \mathcal{P} (A \cup B) = \mathcal{P} (A ) \cup \mathcal{P} (B)$ it follows that $A \cup B \in \mathcal{P} (A)$ or $A \cup B \in \mathcal{P} (B)$.
Thus $A \cup B \subset A$ or $A \cup B \subset B$. Therefore  $B \subset A \cup B \subset A$ or $A \subset A \cup B \subset B$.
A: I have to admit I was confused by the question until I noticed the following. When you think of $\mathcal{P}(A) \cup \mathcal{P}(B)$ watch out. It isn't true that:
$$\mathcal{P}(A) \cup \mathcal{P}(B) = \{\alpha \cup \beta: \alpha \subset A, \beta \subset B\}$$
but rather 
$$\mathcal{P}(A) \cup \mathcal{P}(B) = \{\alpha: \alpha \subset A  \ or \  \alpha \subset B \}.$$
