$P(A − B) − \varnothing ⊆ P(A) − P(B)$ Prove that if $A$ and $B$ be sets. Then $P(A − B) − \{\varnothing\} ⊆ P(A) − P(B)$.
If there was no empty set, I know the rules such as $\{x\}$ is in $P(A)$ implies $x$ in $A$. But I don't know when there is an empty set.
 A: Always type-check when writing set operations (especially involving $P$ or $\varnothing$): When you wrote $P(A-B)-\varnothing$, this means the set containing subsets of $A-B$ but not elements of $\varnothing$. You probably mean the set containing subsets of $A-B$ that is not $\varnothing$, which should be $P(A-B)-\{\varnothing\}$.
Now by definition, we need to check for every $X\in P(A-B)-\{\varnothing\}$, that $X\in P(A)-P(B)$. By definition again, we need to check that if $X\subseteq A-B$ and $X\ne\varnothing$, then $X\subseteq A$ and $X\not\subseteq B$.
To check $X\subseteq A$, note that $X\subseteq A-B\subseteq A$. Then use transitivity of $\subseteq$ to conclude that $X\subseteq A$.
To check that $X\not\subseteq B$, we will use proof by contradiciton. If $X\subseteq B$, then since $X\subseteq A-B$ as well, we know that $X\subseteq (B)\cap(A-B)=\varnothing$. However $X\subseteq\varnothing$ implies $X=\varnothing$, contradicting that $X\ne \varnothing$. Thus by contradiction, we know that $X\not\subseteq B$.
QED

Okay some more formal nonsense: Note that there is the edge case where $A\subseteq B$, or even rather $A\subsetneq B$. The desired inclusion still holds in this case, even though it has completely lost meaning. In particular, $P(A-B)=\{\varnothing\}$, so there would be no such $X\in P(A-B)-\{\varnothing\}$. However since there is no such $X$, it is still valid to say "every such $X$ satisfies ...".
A: $P(A)$ means the set of all subsets of $A$.
$A-B$ means the set of all elements in $A$ that are not in $B$.
Stop and first think about what any set minus $\{\varnothing\}$ would be.
Now, take something, call it $S$, that is in the left hand side. If the left hand side is the empty set, you are finished, so we can assume that there is something in the left hand side. That means $S$ is a subset of the set of elements that are in $A$, but not in $B$. That set is contained in $P(A)$, but it cannot be contained in $P(B)$. If $S$ were contained in $P(B)$, then all the elements of $S$ would be elements of $B$. But, that would mean $S$ would be a subset of $B$. If you are not convinced, take a specific $x$ that is an element of $S$. $S$ cannot be empty because the definition of the left hand side has excluded $\varnothing$. Convince yourself that $x$ has to be an element of $A$ but it cannot be an element of $B$.  Once you are convinced of that, you should see that the entire set $S$ is contained in $P(A)$, but not in $P(B)$.
A: Element chase.
Let $X \in P(A-B) - \emptyset$.  Then $X \subset A-B$ and $X \ne \emptyset$.  So $X$ consists entirely of elements within $A$ and has no elements of $B$.
So $X$ is a subset of $A$.  As $X$ has elements that are not in $B$, $X$ is not a subset of $B$. (Note if $X = \emptyset$ we would have $X$ is a subset of $B$ but because $X$ isn't empty it is not.)
So  $X$ is a subset of $A$.  So $X \in P(A)$.  And $X$ is not a subset of $B$ so $X \not \in P(B)$.  So $X\in P(A) -P(B)$ and $ P(A-B) - \emptyset\subset P(A) -P(B)$.
That's it.
....
Note the other direction does not hold.  A subset $K$ of $A$  that is not a subset of $B$ could contain an element that is in $A$ but not $B$, and an element that is in $A$ and $B$.  But then $K$ will contain elements of $B$ so it will not be a subset of $A-B$.
