Find the mistake of the reasoning Let $A,B,C$ and $D$ be arbitrary sets.
\begin{align*}
\mathcal{P}(A)\subseteq\mathcal{P}(B) &\iff (C\in\mathcal{P}(A)\implies C\in\mathcal{P}(B))&\text{by definition of subclass}\\
&\iff (C\subseteq A\implies C\subseteq B)&\text{by definition of power set}\\
&\iff [(D\in C\implies D\in A)\implies (D\in C\implies D\in B)]&\text{by definition of subclass}\\
\therefore\ \mathcal{P}(A)\subseteq\mathcal{P}(B) &\iff [(D\in C\implies D\in A)\implies (D\in C\implies D\in B)]
\end{align*}
Suppose that $\mathcal{P}(A)\subseteq\mathcal{P}(B) \iff A\subseteq B$.

*

*Since $A\subseteq B\iff (D\in A\implies D\in B)$ we have that $$\mathcal{P}(A)\subseteq\mathcal{P}(B) \iff (D\in A\implies D\in B).$$


*and, $$\mathcal{P}(A)\subseteq\mathcal{P}(B) \iff [(D\in C\implies D\in A)\implies (D\in C\implies D\in B)]$$


*Then, $$(D\in A\implies D\in B)\iff [(D\in C\implies D\in A)\implies (D\in C\implies D\in B)]$$


*But this is false, since it is not the case that $$[(D\in C\implies D\in A)\implies (D\in C\implies D\in B)]\implies (D\in A\implies D\in B)$$ https://www.umsu.de/trees/#((P~5Q)~5(P~5R))~5(Q~5R).


*Therefore it is false that $\mathcal{P}(A)\subseteq\mathcal{P}(B) \iff A\subseteq B$.
What is wrong with my reasoning?
Thank you so much.
 A: 
$$[(D\in C\implies D\in A)\implies (D\in C\implies D\in B)]\implies (D\in A\implies D\in B)$$

If you restore the quantifiers, this formula should be :
$$[\forall C ,( \forall D, D\in C\implies D\in A)\implies (\forall D, D\in C\implies D\in B)]\implies (\forall D,  D\in A\implies D\in B)$$
To prove that this implication holds, assume that :
$$\forall C ,( \forall D, D\in C\implies D\in A)\implies (\forall D, D\in C\implies D\in B)$$
Then, since the LHS of the implication with $C$ replaced by $A$ :
$$ \forall D, D\in A\implies D\in A$$
is true, the RHS must follow, and therefore :
$$\forall D, D\in A\implies D\in B$$
which is what we wanted to prove.
A: 

*

*But this is false, since it is not the case that $$[(D\in C\implies D\in A)\implies (D\in C\implies D\in B)]\implies (D\in A\implies D\in B)$$ https://www.umsu.de/trees/#((P~5Q)~5(P~5R))~5(Q~5R).


The claim is actually true. Notice that the left-hand side says
$$[(C\subset A)\implies (C\subset B)],$$
while if $D\in A$, then $\{D\} \subset A$. Using the above, we get $\{D\}\subset B$, meaning $D\in B$.
A: You checked the formula
$$((P\rightarrow Q)\rightarrow (P\rightarrow R))\rightarrow (Q\rightarrow R)$$
using the "Tree Proof Generator"
https://www.umsu.de/trees/#((P~5Q)~5(P~5R))~5(Q~5R)
and got as result that the formula is invalid and
Countermodel:
P:  false
Q:  true
R:  false

But actually you wanted to check
$$[(D\in C\implies D\in A)\implies (D\in C\implies D\in B)]\implies (D\in A\implies D\in B) \tag 1$$
The "Tree Proof Generator" can handle predicates and a 1-ary predicate is a set. So you can translate your original question to an expression that can handled by the Tree Proof Generator. I use a lowercase $d$ instead of the uppercase $D$ to make this expression more readable.
$$( (Cd  \to Ad  ) \to  (Cd  \to Bd )) \to (Ad  \to Bd ) $$
You can evaluate this by
https://www.umsu.de/trees/#((Cd~5Ad)~5(Cd~5Bd))~5(Ad~5Bd)

Again Tree Proof Generator tells you that the statement is invalid and returns
Countermodel:
Domain: { 0 }
d:  0
C:  { }A:  { 0 }
B:  { }

Now one can see what is wrong with your statement. It is false if you choose these especial values for the variables.
$$((0 \in \emptyset \to 0 \in \{0\}) \to (0 \in \emptyset \to 0 \in \emptyset)) \to (0 \in \{0\} \to 0 \in \emptyset)  $$
But that is not what you wanted to claim with $(1)$. The $d$ in all paranthese expressions should not be the same. We need the $\forall$ quantor:
$$( \forall d (Cd  \to Ad  ) \to  \forall d (Cd  \to Bd )) \to  \forall d (Ad  \to Bd ) $$
Note that this is the same as to
$$( \forall x (Cx  \to Ax  ) \to  \forall y (Cy  \to By )) \to  \forall z (Az  \to Bz ) \tag 3$$
We can  put this in the Tree Proof Generator
https://www.umsu.de/trees/#(~6d(Cd~5Ad)~5~6d(Cd~5Bd))~5~6d(Ad~5Bd)

We get still the message that this is an invalid statement and the output
Countermodel:
Domain: { 0 }
C:  { }A:  { 0 }
B:  { }

We note that $(3)$  this is still not what we want to say. We want not restrict to a special $C$, but the left hand side of the main implication should be true for all sets $C$, not only for one $C$. So we finally get this expressen
$$(\forall C( \forall d (Cd  \to Ad  ) \to  \forall d (Cd  \to Bd ))) \to  \forall d (Ad  \to Bd ) $$
This we cannot proof with the Tree Proof Generator because it cannot handle predicates in quantors. But $(4)$ is true as SolubleFish has already proved here.
