What's wrong with this proof about power sets? As an exercise, I was given the task to prove that $$ \mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A \cup B). $$
Wanting to do it in the most rigorous way possible, I used the logical definitions and passages as simple as possible, and this is what I've come up with:
$ x \in \mathcal{P}(A) \cup \mathcal{P}(B) \Leftrightarrow x \in \mathcal{P}(A) \  \lor x \in \mathcal{P}(B) \Leftrightarrow x \subseteq A \ \lor x \subseteq B \Leftrightarrow $ 
$ \Leftrightarrow \forall y (y \in x \Rightarrow y \in A) \ \lor \ \forall y (y\in x \Rightarrow y \in B) \color{red}{ \boldsymbol\Leftrightarrow } \forall y (y\in x \Rightarrow y \in A \ \lor y \in B) \Leftrightarrow $
$ \Leftrightarrow \forall y (y\in x \Rightarrow y \in A \cup B) \Leftrightarrow x \subseteq A \cup B \Leftrightarrow x \in \mathcal{P}(A \cup B).$
All of this seems to prove that $ \forall x(x \in \mathcal{P}(A) \cup \mathcal{P}(B) \Leftrightarrow x \in \mathcal{P}(A \cup B)) $ or equivalently that $ \mathcal{P}(A) \cup \mathcal{P}(B) = \mathcal{P}(A \cup B) $, so I asked myself "Why does the exercise only ask to prove half of it?", and I rapidly realized that, for example, $ A \cup B $ is always an element of $\mathcal{P}(A \cup B) $, but it's not generally a subset of $A$ or $B$, so it's not necessarily an element in $\mathcal{P}(A) \cup \mathcal{P}(B)$.
So I think there must be a mistake in my procedure. I highlighted in red the passage I'm not confident with, but I don't understand why it should be wrong.
 A: It is not true that :

$\forall x (Px \lor Qx) \leftrightarrow (\forall x Px \lor \forall x Qx)$.

Consider the counter-example: "every (natural) number is either odd or even".
This is : $\forall n (\text {Odd}(n) \lor \text {Even}(n))$.
But it is not true that : "either all numbers are odd or all numbers are even".

Thus the "central" step in your proof must be :

$(∀y(y∈x ⇒ y∈A) ∨ ∀y(y∈x ⇒ y∈B)) ⇒ ∀y(y∈x ⇒ (y∈A ∨y∈B))$.

As you corectly say, the fact that $\mathcal P(A∪B) \nsubseteq \mathcal P(A) ∪ \mathcal P(B)$ is another counter-example for the above equivalence.
A: This is an example where the notation obscures the content. "Rigorous" means "utterly convincing", not "expressed using the smallest alphabet of symbols".
You're right that your red step is wrong. The problem is that you've squashed the two independent "for all $y$" expressions together.
Consider the following plain-English statements: "for all dogs, if the dog is red then it's big", and "for all dogs, if the dog is red then it's small". I can express them more Englishily as "all red dogs are big" and "all red dogs are small" respectively.
You're asserting that the logical-Or of those two is equivalent to "all red dogs are either big or small", which is clearly a false equivalence: there are both big and small dogs, so "all red dogs are big or small" can hold even if "all red dogs are big" and "all red dogs are small" are both false.
