My problem:
Prove that for all sets $A$ and $B$, $P(A)\cap P(B) = P(A\cap B)$.
I have attempted to solve is as follows. I would like some confirmation as to whether or not my proof is valid.
My Solution:
Let $X \in P(A \cap B)$.
Then every element of $X$ is an element of $A$ and $B$, thus $X$ is also in $P(A)$ and $P(B) \implies X \in P(A)\cap P(B)$.
Now, let $Y \in P(A)\cap P(B)$.
Then $Y \in P(A) $ and $Y \in P(B)$. Therefore each element of $Y$ is an element of $A$ and $B$. Thus, each element of $Y$ is in $A\cap B \implies Y \in P(A\cap B)$.
$X$ and $Y$ are arbitrary. Thus, any set in $P(A \cap B)$ is in $P(A)\cap P(B)$ and vice-versa.
From this, we are able to conclude the two sets have an identical composition and thus equal.
I'd make it a bit more explicit: If $X \in P(A \cap B)$ then $X \subseteq A \cap B\subseteq A$ which implies that $X \subseteq A$, so $X \in P(A)$, and also that $X \subseteq B$ (from $A \cap B \subseteq B$) so that $X \in P(B)$, and then we can conclude $X \in P(A) \cap P(B)$.
If $X \in P(A)$ and $X \in P(B)$ then $X \subseteq A$ and $X \subseteq B$, so that $X \subseteq A \cap B$ and thus by definition, $X \in P(A \cap B)$.
Showing two inclusions as you do is fine.