Q:

Is my proof of if $B \subseteq A$ then $P(B) \setminus P(A) = \emptyset$ correct?

My try:
Suppose $B \subseteq A$.
Further suppose $X \in P(B) \setminus P(A)$. By the definition of the "$\setminus$" operator, we know $X \in P(B)$ and $X \notin P(A)$. By the properties of powerset, we know $X \subseteq B$ and by the properties that we proved, we know $X \subseteq A$. Hence, by the definition of powerset, we got that $X \in P(A)$. But that's a contradiction, because $X \in P(B)$ and $X \notin P(A)$.
Hence, by absurd, we got that $P(A) \setminus P(B) = \emptyset$

A:

Your proof by contradiction looks fine to me (once the formatting problems were fixed). I would make one adjustment. Instead of saying:

By the properties of powerset, we know $X \subseteq B$ and by the properties that we proved, we know $X \subseteq A$.

I would word this as:

By the definition of a power set, we know that $X \subseteq B$. But since our initial assumption was that $B \subseteq A$, it follows by the transitivity of the $\subseteq$ relation that $X \subseteq A$.

