Proof Verification: If $P(A \cup B) = P(A) \cup P(B)$, then either $A \subset B$ or $B \subset A$

Question

I have an exercise stating

Prove that if $P(A \cup B) = P(A) \cup P(B)$, then either $A \subset B$ or $B \subset A$. (Where $P(S)$ denotes the power set of $S$.

It gives the hint to prove the contrapositive. My proof is:

The contrapositive of the statement is $(A \not\subset B) \land (B \not\subset A) \implies P(A\cup B) \neq P(A)\cup P(B)$. Given some $a \in A$ and $b\in B$ such that $a\notin B$ and $b\notin A$, it is true that $(A\not\subset B)\land (B\not\subset A)$. From this we get $\{a,b \} \notin P(A)$ and $\{ a,b\}\notin P(B)$, and therefore $\{ a,b\} \notin P(A) \cup P(B)$. It is also true that $a\in A\cup B$ and $b \in A\cup B$, so $\{ a,b\}\in P(A \cup B)$. Therefore, $P(A \cup B) \neq P(A) \cup P(B)$

Is this proof correct, and if so is there anything that would be better expressed differently.

Answer 1

The proof idea is certainly correct, and its execution is almost correct. Just one small issue right at the start:

Given some $a \in A$ and $b\in B$ such that $a\notin B$ and $b\notin A$, it is true that $(A\not\subset B)\land (B\not\subset A)$ ....

You need to go just the other way around, i.e. you need to say:

Assume that $(A\not\subset B)\land (B\not\subset A)$. Then it must be the case that there is some $a \in A$ and $b\in B$ such that $a\notin B$ and $b\notin A$ ...

Why? Because you are going from $(A\not\subset B)\land (B\not\subset A)$ to $P(A\cup B) \neq P(A)\cup P(B)$, so your very start should be $(A\not\subset B)\land (B\not\subset A)$