Seems obvious but I doubt how to show it step by step... I could prove the reverse statement (if $A\subseteq B$ or $B\subseteq A$ then $\mathcal{P}(A)\cup\mathcal{P}(B)=\mathcal{P}(A\cup B)$), but it doesn't help.
I saw a hint that it's easy to prove by negation, but I cant see how...
Thanks for help at advance!
Edit: $\mathcal{P}(X)$ notes the powerset of $X$
Hint:
If $A \not\subset B$ nor $B \not\subset A$ then $\exists a \in A \setminus B, b \in B \setminus A$. Think about some element of $\mathcal{P}(A \cup B)$ that couldn't possibly belong to $\mathcal{P}(A) \cup \mathcal{P}(B)$.
Edit:
thinking about your comment it could actually constitute a proof (just for the finite case)
Clearly $|\mathcal{P}(A)| = 2^{|A|}, |\mathcal{P}(B)| = 2^{|B|}$.
$|\mathcal{P}(A) \cup \mathcal{P}(B)| = 2^{|A|} + 2^{|B|} - |\mathcal{P}(A) \cap \mathcal{P}(B)| = 2^{|A|} + 2^{|B|} - |\mathcal{P}(A \cap B)| = 2^{|A|} + 2^{|B|} - 2^{|A \cap B|}$ $|\mathcal{P}(A \cup B)| = 2^{|A| + |B| - |A \cap B|}$
Now if $A \not\subset B$ and $B \not\subset A$ then $|A \cap B| < \min \{|A|, |B|\}$. This way, $|\mathcal{P}(A) \cup \mathcal{P}(B)| = 2^{|A|} + 2^{|B|} - 2^{|A \cap B|}$ has, at least, three $1$s in its binary representation, while $|\mathcal{P}(A \cup B)| = 2^{|A| + |B| - |A \cap B|}$ has just one.
