This question already has an answer here:
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 $F$.
If $\mathcal P(A)\cup \mathcal P(B)=\mathcal P(A\cup B)$, then $A\cup B\in \mathcal P(A)\cup \mathcal P(B)$, so either $A\cup B\in \mathcal P(A)$ or $A\cup B\in \mathcal P(B)$. Thus either $A\cup B\subseteq A$ or $A\cup B\subseteq B$, so…
You’re right: $\wp(A)\cup\wp(B)\ne\wp(A\cup B)$ doesn’t obviously give you much to work with; that’s often the case with not equals statements. When that happens, it’s always worth taking a look at the contrapositive: does negating the desired conclusion give you more to work with? Here that gives you the hypothesis that $A\nsubseteq B\nsubseteq A$; not a subset isn’t necessarily much better than not equals, but it’s a statement about simpler objects, and when re-expressed in terms of elements of $A$ and $B$, it turns out to do the trick. That’s Arthur Fischer’s excellent hint.
The next step, by the way, would be to try a proof by contradiction: that gives you the ‘largest’ possible hypothesis to work with.
However, the original hypothesis actually does more than may at first appear, if you think about it right. It’s pretty obvious that $\wp(A)\cup\wp(B)\subseteq\wp(A\cup B)$ always holds, so $\wp(A)\cup\wp(B)\ne\wp(A\cup B)$ is really saying that $\wp(A)\cup\wp(B)\subsetneqq\wp(A\cup B)$, i.e., that $A\cup B$ has some subset $S$ that is not a subset of $A$ and not a subset of $B$. Draw a Venn diagram, and you’ll probably see immediately why that entails $A\nsubseteq B\nsubseteq A$.