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: There is a simple schematic for conditional proofs $P\implies Q$. However, keep in mind, at no point, you can assume anything about the consequent. 
You suppose $P$, and show $Q$.
In this case, Suppose: $$B\subseteq A$$
Now, suppose:$$X\subseteq A$$
If $X\subseteq B$, it follows from out assumption, $B\subseteq A$, that $X\subseteq A$.
Therefore, $\forall X\subseteq B$, it follows that $X\subseteq A$.  
By definition of power set: $\forall X\subseteq B \implies X\in P(B)$, by the proof above it implies $X\in P(A)$ aswell. 
Thus, if all the elements in $P(B)$ are also in $P(A)$, then that implies $P(B)\setminus P(A)=\emptyset$.
Every consequent follows from the antecedent. Which, in turn, follows from the initial assumption. Clear and concise.
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 $X \subseteq A$.

A: I hope this proof helps:
Prove: $A\subseteq B$ iff $A\setminus B = \emptyset$
Proof:
$\Rightarrow$ Suppose $A\subseteq B$ and $x\in (A\setminus B)$. Then $x\in A$ and $x\in B$ since $A\subseteq B$. So, $x\in A\cap B$ which implies $A\cap B\subseteq B$. Note we can write $$A = (A\cap B)\cup (A\setminus B)$$
then clearly we have $A\setminus B = \emptyset$.
$\Leftarrow$ Conversely, suppose $A\setminus B = \emptyset$, if $x\notin B$ then $x\in A\setminus B = \emptyset$, so $x\in B$ thus $A\subseteq B$.
