This is more interesting than the usual stuff.
Prove P(A) ⊆ P(B) ↔ A ⊆ B
It's an if and only if, we'll do → first. Show P(A) ⊆ P(B) → A ⊆ B
Let's assume it's false; We have A⊄B and P(A) ⊆ P(B).
A⊄B means there's an element a∈A such that a∉B. So the singleton subset {a} must be in the powerset of A,
{a} ∈ P(A)
Since P(A) ⊆ P(B) it follows {a} ⊆ P(B)
But that can't possibly be true because a∉B. We found our contradication, so we must have P(A) ⊆ P(B) → A ⊆ B.
← next. Show A ⊆ B → P(A) ⊆ P(B)
Again we'll assume it's false, assume A ⊆ B and P(A) ⊄ P(B)
Since P(A) ⊄ P(B) there must be at least one set S∈P(A) and S∉P(B).
S can't be the empty set because the empty set is in every power set; it's a subset of every set. So S has at least one element, and all the elements must have come from A.
S ≠ ∅
a∈S → a∈A
Since S∉P(B) that means there exists at least one element, let's call it a, such that a is in S but not in B.
a∈S and a∉B
So a∈A and a∉B, contradicting A ⊆ B. We conclude A ⊆ B → P(A) ⊆ P(B).
That proves both ways.