Prove that $\mathcal{P}(A)⊆ \mathcal{P}(B)$ if and only if $A⊆B$. [duplicate]
Your argument is fine, if a little more roundabout than necessary: both directions can be done easily with direct proofs.
Suppose first that $\wp(A)\subseteq\wp(B)$. $A\subseteq A$, so $A\in\wp(A)\subseteq\wp(B)$, so $A\in\wp(B)$, and hence $A\subseteq B$.
Now suppose that $A\subseteq B$. Then for any $X\in\wp(A)$ we have $X\subseteq A\subseteq B$, so $X\subseteq B$, and therefore $X\in\wp(B)$. Thus, $\wp(A)\subseteq\wp(B)$.
I know you were asking for a critique of your original proof. But since I find text-based proofs (like yours, and the one from another answer) more difficult to read than symbolic ones, let me answer by presenting the way in which I would write a proof for this.
Proof. For all sets A and B,
$$ \begin{array}{ll} & \mathcal{P}(A) ⊆ \mathcal{P}(B) \\ \equiv & \;\;\;\text{"definition of ⊆"} \\ & \langle \forall V : V \in \mathcal{P}(A) : V \in \mathcal{P}(B) \rangle \\ \equiv & \;\;\;\text{"definition of $\mathcal{P}$, twice"} \\ & \langle \forall V : V ⊆ A : V ⊆ B \rangle \;\;\;\;\;\;\;\;\;\;\;\; (*) \\ \equiv & \;\;\;\text{"see below"} \\ & A ⊆ B \\ \end{array} $$
which proves this theorem. The forward direction in the last step is easily proven:
$$ \begin{array}{ll} & \langle \forall V : V ⊆ A : V ⊆ B \rangle \\ \Rightarrow & \;\;\;\text{"choose $V := A$; $A ⊆ A$"} \\ & A ⊆ B \\ \end{array} $$
For the other direction, assuming $A ⊆ B$ we prove $(*)$ as follows: for every $V$,
$$ \begin{array}{ll} & V ⊆ A \\ \Rightarrow & \;\;\;\text{"A ⊆ B, and ⊆ is transitive"} \\ & V ⊆ B \\ \end{array} $$
Note. This is the style of proof writing used in, e.g., A Logical Approach to Discrete Math by Gries and Schneider; it was originally designed by Dijkstra and Feijen, and is discussed in chapter "On our proof format" of Dijkstra and Scholten, Predicate calculus and program semantics, and (to provide a more accessible source) near the end of EWD1300.
The nice thing about this style is that the $\text{"hints"}$ show very clearly which properties are used: apart from predicate logic and the definitions of $\mathcal{P}$ and $⊆$, the above proof only uses the fact that $⊆$ is reflexive and transitive.