Prove that $\mathscr P(A)\cap\mathscr P(B)=\mathscr P(A\cap B)$.
Consider the set $X=\{x\}$ with $X\in\mathscr P(A)\cap\mathscr P(B)$. This means that $X\in\mathscr P(A)$ and $X\in\mathscr P(B)$ and so $x\in A$ and $x\in B$.Then by the definition of intersection $x\in(A\cap B)$. But then by the definition of the powerset, we have $X\in\mathscr P(A\cap B)$ which means $\mathscr P(A)\cap\mathscr P(B)\subseteq\mathscr P(A\cap B)$.
Now consider the set $Y=\{y\}$ where $Y\in\mathscr P(A\cap B)$. Then we have $y\in A\cap B$. THis means $y\in A$ and $y\in B$. Therefore we must have $Y\in\mathscr P(A)$ and $Y\in\mathscr P(B)$ and so by the definition of intersection we have $Y\in\mathscr P(A)\cap\mathscr P(B)$ and thus $\mathscr P(A\cap B)\subseteq\mathscr P(A)\cap\mathscr P(B)$.
By the above arguments we can see that $\mathscr P(A)\cap\mathscr P(B)=\mathscr P(A\cap B)$.
$\blacksquare$
My concern is that this proof is invalid because I followed the exact same argument in part 2 as in part 1 but in reverse. Is this a problem or does my proof work? Thanks!
Your proof is invalid because you've only shown the sets you claim are equal contain the same singletons. You should use the general observation$$\begin{align}X\in\mathscr{P}(A)\cap\mathscr{P}(B)&\iff X\in\mathscr{P}(A)\land X\in\mathscr{P}(B)\\&\iff X\subseteq A\land X\subseteq B\\&\iff\forall x\in X(x\in A)\land\forall x\in X(x\in B)\\&\iff\forall x\in X(x\in A\cap B)\\&\iff X\subseteq A\cap B\\&\iff X\in\mathscr{P}(A\cap B).\end{align}$$