Why is $\mathcal{P}(A\cup B) \not\subset \mathcal{P}(A)\cup\mathcal{P}(B)$?

I had a question in my 11th grade Mathematics book:

Prove that $\mathcal{P}(A\cup B) \not\subset \mathcal{P}(A)\cup\mathcal{P}(B)$.

I did a proof for why $\mathcal{P}(A\cup B) \subset \mathcal{P}(A)\cup P(B)$

Which step in the following proof is wrong?

Proof

Let $y$ be an arbitrary element of a set $Y$
Let $Y$ be an element of $\mathcal{P}(A\cup B)$
So, $Y \subset (A\cup B)$
So, $y \in (A\cup B)$
So, $y \in A$ or $y ∈ B$
So, $Y \subset A$ or $Y \subset B$
So, $Y \in \mathcal{P}(A)$ or $Y ∈ \mathcal{P}(B)$
So, $Y \in \mathcal{P}(A)\cup \mathcal{P}(B)$
And thus: every element of $\mathcal{P}(A\cup B)$ is an element of $\mathcal{P}(A)\cup P(B)$
So, $\mathcal{P}(A\cup B) \subset \mathcal{P}(A)\cup P(B)$

I suspect a mistake in the 6th line but I just wanted confirmation.

Thank You...

The 6th line is wrong. What you have shown is that an arbitrary $y\in Y$ is in $A$ or is in $B$ but this doesn't mean ALL $y\in Y$ satisfy this for same set $A$ or $B$. You might have some $y\in Y$ who are in $A$ and some $y'\in Y$ who are in $B$. Then, $Y$ is not an element of $P(A)$ or of $P(B)$ but is an element of of $P(A\cup B)$.