The following proof is given to show that $\mathcal P(A) \cup \mathcal P(B) \subseteq \mathcal P(A\cup B)$ ($\mathcal P$ is the power set):
Let $X \in (\mathcal{P}(A) \cup \mathcal{P}(B))$, by definition of union: $$X \in \mathcal{P}(A) \lor X \in \mathcal{P}(B)$$ $\implies$By definition of power set $$X \subseteq A \lor X \subseteq B$$ $\implies$By union transitivity$$X \subseteq A\cup B$$ $\implies$By definition of power set $$X \in \mathcal{P}(A \cup B)$$ Then $\mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A \cup B) \blacksquare$
Now, I thought I could apply the opposite logic to show that $\mathcal P(A) \cup \mathcal P(B) \supseteq \mathcal P(A\cup B)$, and thus prove that $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$:
Let $X \in \mathcal{P}(A \cup B)$, By definition of power set: $$X \subseteq A\cup B$$ $\implies$ By union transitivity: $$X \subseteq A \lor X \subseteq B$$ $\implies$ By definition of powerset: $$X \in \mathcal{P}(A) \lor X \in \mathcal{P}(B)$$ $\implies$ by definition of union: $$X \in (\mathcal{P}(A) \cup \mathcal{P}(B))$$ Then $\mathcal {P}(A \cup B) \subseteq \mathcal{P}(A) \cup \mathcal{P}(B) \blacksquare$
Multiple posts here prove that for any sets $A$ or $B$, if $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$ then either $A \subseteq B$ or $B \subseteq A$, so clearly my opposite direction proof can't be true.
What I'd like to understand is which step is based on a false assumtion/implication?
$X \subseteq A \cup B$ doesn't imply $X \subseteq A$ or $X \subseteq B$.
For example let $A=\{1\}$ and $B=\{2\}$ and $X = A \cup B$.
$X$ is neither subset of $A$ nor subset of $B$.