Power set of Set differences 
Assume that $\mathcal P(A-B)= \mathcal P(A)$. Prove that $A\cap B = \varnothing$.

What I did:
I tried proving this directly and I got stuck.
Let $X$ represent a nonempty set, and let $X\in\mathcal P(A-B)$.
By definition of power set and set difference:
$X\subseteq A$ and $X\nsubseteq B$.
By definition of a subset and subset negation:
Let $y$ be an arbitrary element of $X$ such that:
$(\forall y)(y \in X \rightarrow y \in A)\land (\exists y)(y \in X \land y \nsubseteq B)$ 
This is where I get stuck, how do I confirm I have no common elements with the fact that there is at least one element they don't share?
I also tried proving the contrapositive by assuming that the intersection is not disjoint, where I let $X$ be a subset in the intersection then $X$ must belong to both $A$ and $B$ and that's where I got stuck.
I apologize for bad formatting first time poster 

Think only about the set $A$ itself:
$A \in \mathcal P(A) $ is a subset of $A\setminus B \subseteq A$ iff $A\setminus B = A$ iff $A \cap B=\emptyset$
