This is the problem I am trying to answer:
Let A and B be sets. Prove that $A ⊆ B ⇔ P(P(A)) ⊆ P(P(B))$
This is my proof:
Assume $A ⊆ B$. If $x ⊆ A$, then $x ⊆ B$, so $x ∈ P(A)$ and $x ∈ P(B)$. Thus, $P(A) ⊆ P(B)$. Now, if $x ⊆ P(A)$, then $x ⊆ P(B)$, so $x ∈ P(P(A))$ and $x ∈ P(P(B)$. Therefore, $P(P(A)) ⊆ P(P(B))$ and $A ⊆ B ⇒ P(P(A)) ⊆ P(P(B))$. Now assume $P(P(A)) ⊆ P(P(B))$. If $x ∈ P(P(A))$, then $x ∈ P(P(B))$, so $x ⊆ P(A)$ and $x ⊆ P(B)$, so $P(A) ⊆ P(B)$. Now if $x ∈ P(A)$, then $x ∈ P(B)$, so $x ⊆ A$ and $x ⊆ B$. Thus $A ⊆ B$ and $P(P(A)) ⊆ P(P(B)) ⇒ A ⊆ B$. Hence $A ⊆ B ⇔ P(P(A)) ⊆ P(P(B))$.
I think I might not have a clear understanding of subsets vs elements and may have been terribly wrong in my proof, but have given it my best shot. Can someone check if this proof makes logical sense and if/where I went wrong?
Your proof is basically good, though the converse direction could be made clearer.
Two observations:
You can use different (kind of) letters to denote elements, subsets and subsets of subsets, especially if these confuse you.
Actually we have $A\subseteq B\iff P(A) \subseteq P(B)$, and you can apply it twice.
An easy proof for the converse:
If $P(A)\subseteq P(B)$, then in particular $A\in P(A)\subseteq P(B)$ meaning directly that $A\subseteq B$.