The power set of the intersection of two sets equals the intersection of the power sets of each set

"Again, take any subset $X \subseteq A \cap B$."

You might want to change the wording here. "Take any set $X$ such that $X \subseteq A$ and $X \subseteq B$. Then $X \subseteq A \cap B$."

As written it sounds like you're assuming what you want to prove. Otherwise good...

For the first half, you want to prove that any member of $\mathcal{P}(A \cap B)$ is a member of $\mathcal{P}(A) \cap \mathcal{P}(B)$. So the proof should start "Suppose $X \in \mathcal{P}(A \cap B)$". Then you derive $X \subseteq A \cap B$, and the rest goes as you state. (Except, don't use "means" for "implies".)

For the second half, you aren't going to use "just the same arguments" or you'd end up showing that $\mathcal{P}(A \cap B) \subseteq \mathcal{P}(A \cap B)$ which isn't what you meant!

$ \newcommand{pow}[1]{\mathcal P(#1)} $For comparison, here is another way to prove this: start with the most complex side, investigate which elements are in that set by expanding the definitions, and then try to simplify. So we calculate, for any $\;V\;$, \begin{align} & V \in \pow A \cap \pow B \\ \equiv & \qquad \text{"definition of $\;\cap\;$"} \\ & V \in \pow A \;\land\; V \in \pow B \\ \equiv & \qquad \text{"definition of $\;\pow{\quad}\;$, twice"} \\ & V \subseteq A \;\land V \subseteq B \\ \equiv & \qquad \text{"set theory: property of $\;\cap\;$"} \\ & V \subseteq A \cap B \\ \equiv & \qquad \text{"definition of $\;\pow{\quad}\;$"} \\ & V \in \pow{A \cap B} \\ \end{align} By set extensionality, this proves $\;\pow A \cap \pow B \;=\; \pow{A \cap B}\;$.