$A$ is a subset of $B$ if and only if $P(A) \subset P(B)$ I had to prove the following for a trial calculus exam:
$A\subset B$ if and only if $P(A) \subset P(B)$ where $P(A)$ is the set of all subsets of $A$.
Can someone tell me if my approach is correct and please give the correct proof otherwise?
$PROOF$:
$\Big(\Longrightarrow\Big)$ assume $A\subset B$ is true. Then $\forall$ $a\in A$, $a\in B$
Then for $\forall$ A, the elements $a_1, a_2,$ ... , $a_n$ in A are also in B.
Hence $P(A)\subset P(B)$
$\Big(\Longleftarrow\Big) $ assume $P(A) \subset P(B)$ is true. We prove this by contradiction so assume $A\not\subset B$
Then there is a set $A$ with an element $a$ in it, $a\notin$ B.
Hence $P(A) \not\subset P(B)$
But we assumed $P(A) = P(B)$ is true.
We reached a contradiction.
Hence if $P(A) = P(B)$ then $A\subset B$.

I proved it both sides now, please improve me if I did something wrong :-)
 A: Neither direction seems to be valid. Here is the correct approach.  
$(\Rightarrow)$  Assume $A \subseteq B$.  Then for any $C \in P(A)$, we have $C \subseteq A \subseteq B$.  Hence, $C \in P(B)$.
$(\Leftarrow)$  Assume $P(A) \subseteq P(B)$.   Since $A \in P(A)$, $A \in P(B)$, meaning $A \subseteq B$.
A: $(\Rightarrow)$ Given any $x\in P(A)$ then $x\subset A$. So, by hypothesis, $x\subset B$ and so $x\in P(B)$.
The counter part is easier, as cited by @Asaf, below.
A: Hint: $A\in P(A)$. So you don't need proof by contradiction for the second part.
The first part, by enumerating elements as you have, assumes that there are finitely many $a_i$. That's sloppy notation.
Rather, assume $A\subseteq B$.  
If $X\in P(A)$, then $X\subseteq A$, hence $$\forall x\in X: x\in A$$ But $x\in A\implies x\in B$. So we conclude: $$\forall x\in X:x\in B$$ This means that $X\subseteq B$, and hence $X\in P(B)$. So we have shown that if $X\in P(A)$ then $X\in P(B)$, therefore $P(A)\subseteq P(B)$.
A: This is basically correct, but you may be asked to be more precise. You shouldn't use the fact that $a\in A$ is not in $B$ to say $P(A)$ isn't equal to $P(B)$. Instead, say that $P(A)$ isn't contained in $P(B)$. 
