Validating a proof for necessary and sufficient conditions on sets A & B such that $P(A) \cup P(B) = P(A\cup B)$ 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 $B\subseteq A.$
 A: The first part is good.
When you say

Therefore, $P(B) \subseteq P(A) \cup P(B).$

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.
A: Solution: $A \subseteq (A\cup B) \implies P(A) \subseteq P(A\cup B) ; B \subseteq (A \cup B) \implies P(B) \subseteq P(A \cup B)$. 
So, $P(A) \cup P(B) \subseteq P(A\cup B)$ always. 
Conversely, lets claim that $P(A\cup B) \subseteq P(A) \cup P(B)$ if and only if $A\subseteq B$ or $B\subseteq A$. 
Suppose $P(A\cup B) \subseteq P(A) \cup P(B)$. 
If $A\nsubseteq B$ and $\ B \nsubseteq A$. Then there exists some $x\in A$ such that $x \notin B$ and $y\in B$ such that $y\notin A$. Let S= {x,y}
Then it is clear that $S\subseteq (A\cup B) \implies S\in P(A\cup B) \subseteq P(A) \cup P(B)$.
Also, $S\nsubseteq A$ and $S \nsubseteq B$ therefore, $x\in S$, but $x\notin B$ and $y\in S$, but $y\notin A$. Hence $S\notin P(A)$ and $S \notin P(B) \implies S\notin P(A) \cup P(B)$ which is a contradiction because $S\in P(A\cup B) \subseteq P(A) \cup P(B)$.
So, $A\subseteq B$ or $B\subseteq A$. 
If $A\subseteq B$ or $B\subseteq A \implies A\cup B\subseteq B$ or $A\cup B\subset A$.
So, $P(A\cup B) \subseteq P(B)$ or $P(A\cup B) \subseteq P(A) \implies P(A\cup B) \subseteq P(A) \cup P(B)$. 
Thus, $P(A\cup B) = P(A)\cup P(B)$ if and only if $A \subseteq B$ or $B\subseteq A$.
