We will prove both of the expressions $A\subset B$ and $B\subset A$. The first is more difficult, but it follows from $$A_n\subset B_{n+1},$$which we demonstrate below.
First, though, let us verify that $A_n\in B_n$ through induction. Clearly, $A_0\in B_0$, and if $A_n\in B_n$, then $A_{n+1}=P(A_n)\in B_{n+1}$.
Next, we define a sequence $C_n$ by $$C_0=\emptyset\qquad C_{n+1}=\{C_n\}.$$ For $n>0$, the subsets of $A_n$ are classified by those do not contain $C_n$ and those which are equal to a subset of $A_{n-1}$ plus the element $C_n$.
To show that $A_n\subset B_{n+1}$, we begin with $A_0\subset B_0\subset B_1$, establishing the base case of another proof by induction. Assume that $A_n\subset B_{n+1}$. Let $S\subset A_n$, i.e. an element of $A_{n+1}$. If $S$ is empty, then it is an element of $B_0\subset B_{n+2}$. If the nonempty set $S\subset A_{n-1}$, then by hypothesis $S\subset B_{n}\subset B_{n+2}$. Otherwise $C_{n+1}\in S$. There is a subset $T\subset A_{n-1}$ such that $S=A_n\setminus T.$ Since $A_n$ and $T$ are both elements of $B_{n+1}$, it follows that their difference $S\in B_{n+2}$, concluding the proof.
Therefore, $A\subset B$. The other direction is implied by the proposition $$B_n\subset A_{n+1}.$$ Obviously, $B_0\subset A_1$, so assume the induction hypothesis that $B_n\subset A_{n+1}$. If $X$ is an element of $B_n$, then it is a subset of $A_n$. So, its power set $P(X)$ is a a subset of $P(A_n)=A_{n+1}$. That is $x\in A_{n+2}$. If $X,Y\in B_n$, then $$X-Y\subset X\subset A_n.$$ Hence, the difference is an element of $A_{n+1}\subset A_{n+2}$.