HINT: $A\in\mathcal P(A)$, so $A\in\mathcal P(A-B)$. What can you say about all the elements in $\mathcal P(A-B)$ and $B$?
Now that we covered the approach to the correct answer, let me give you a short critique about your efforts.
If $X\in\mathcal P(A-B)$ it is not just that $X\nsubseteq B$, it is that $X\cap B=\varnothing$, this is a much stronger fact, and we do use it here, as seen above.
In the explicit writing that $X\subseteq A$ and $X\nsubseteq B$, you say that there is some $y\in X$ such that $y\nsubseteq B$. That might be a typo, and you meant $y\notin B$, but nonetheless you should be wary about mixing $\in$ and $\subseteq$.
As before, you write in the last attempt that you try to take $X$ in the intersection of $A$ and $B$, then it is a subset of both, this is also not true. It is an element of both. And a subset of the intersection is a subset of both $\mathcal P(A)$ and $\mathcal P(B)$. But that's mainly what you can say about that.
You have a few mistakes:
$\emptyset \subseteq A$ is $\textbf{true}$ as every single element of $\emptyset$ (there are none) are included in $A$.
$\emptyset \subset A$ is $\textbf{true}$ for the same reasoning as above and because $\emptyset \neq A$
$\emptyset \subseteq \{\emptyset\}$ is $\textbf{true}$ for above reasons
$\emptyset \subseteq \mathcal{P}(A)$ for above reasons
$\{\emptyset\} \subseteq \mathcal{P}(A)$ is $\textbf{true}$ because every element of $\{\emptyset\}$ is contained in $\mathcal{P}(A)$ as $\emptyset \in \mathcal{P}(A)$.
Remember that the power set is a set of sets, i.e if $C = \{a,b,c\}$ then
$\mathcal{P}(C) = \{\emptyset,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}\}$
It seems like you have misunderstood the subset relation $\subseteq$. We write $A \subseteq B$ whenever every element of $A$ is included in $B$. If $A$ has no elements, this is trivially true.
Best Answer
It is not clear where exactly you're running into trouble, but from the fact that you can generate examples, it seems that you at least understand the statement. With that said, I suspect the following hint might be useful.
To prove that $\mathcal{P}(A\setminus B ) \subseteq (\mathcal{P}(A)\setminus\mathcal{P}(B)) \cup \ \{\varnothing\}$, it suffices to show that for any $S \in \mathcal P(A \setminus B)$ with $S \neq \varnothing$, $S \in \mathcal P(A) \setminus \mathcal P(B)$. In other words: show that if $S \subseteq A \setminus B$ with $S \neq \varnothing$, then it holds that $S \subseteq A$ and it is not the case that $S \subseteq B$. It is useful to note that if $A \setminus B$ is non-empty, then there exists at least one element $x$ such that $x \in A$ but $x \notin B$.