My proof that if $P(A) \subseteq P(B)$, then $A \subseteq B$ I'm not sure if my proof is sound.
Here it is:

Assume that $P(A) \subseteq P(B)$, so any subset C of A is also a subset of B. Therefore, any element in C is also an element of A, and by the same reasoning is an element of B. Thus $A \subseteq B$, as all the elements in A are proven to be elements in B.

I believe that I've covered that every single element in A must be an element in B, but for some reason I feel doubtful. Is my proof correct?
 A: Half-line proof, without using elements:
$A\in\mathcal P(A)$, hence $A\in\mathcal P(B)$, which means $A\subseteq B$.
A: Yes, you can sort of do as you did and use the fact that $$x\in A\implies \{x\}\in P(A)\stackrel{\text{hypothesis}}{\implies} \{x\}\in P(B)\implies x\in B$$
But you can also directly observe that $$A\subseteq B\stackrel{\text{by def.}}{\iff} A\in P(B)$$ and use $A\in P(A)$ to conclude.
A: "Assume that P(A)⊆P(B), so any subset C of A is also a subset of B." True.
"Therefore, any element in C is also an element of A," 
Here you are missing a quantifier. This is true for all C, but which C are you choosing? In order to prove $A\subseteq B$, you need to start with an arbitrary element of $A$ and show it is in $B$. You've started with an arbitrary element of an unspecified set $C$.
Here's a pretty easy proof. Since $P(A)\subseteq P(B)$ and $A\in P(A)$, that means $A\in P(B)$, so $A\subseteq B$. 
A: Yes the idea is right. I might phrase it differently thought, without the set C. 

Let $a\in A$, now $\{a\}\in P(A)\subseteq P(B)$, and thus $\{a\}\in P(B)$, and hence $a\in B$. 

