Is the solution correct for this?
Problem: Find necessary and sufficient conditions on sets $A$ and $B$ such that the following relation holds among the power sets: $P(A)\cup P(B) = P(A\cup B)$.
Answer: Suppose first that $A\nsubseteq B$ and $B\nsubseteq A$. Then there exists $a \in A \setminus B$ and $b \in B \setminus A$. The subset, {a,b} of $A \cup B$ is not a subset of $A$, nor is it a subset of $B$. Thus, we do not have equality.
Suppose that $A\subseteq B$, then $A \cup B = B$. So, $P(B) \subseteq P(A) \cup P(B)$ by the definition of power sets. Moreover, all subsets of $A$ are subset of $B$, thus you arrive to $P(A) \cup P(B) = P(A \cup B)$. It is the same if $B \subseteq A$.
Thus, the condition is that $A \subseteq B$ or $B\subseteq A.$
Best Answer
The first part is good.
When you say
there is nothing "therefore" about it. $X\subseteq Y\cup X$ is basically the definition of $\cup$.
I would prefer to see some justification for why $$ A\subseteq B\implies P(A)\subseteq P(B) $$ It doesn't have to be long, as it's a quote obvious fact, but just a couple of words acknowledging that it isn't trivial. It is, after all, the real lynchpin of the entire argument.