Restating my question from the title
If the property P is true for all $x\in A$ and P is true for all $y\in B$, is $P(A \cup B)$ true?
I argue that the indeed $P(A \cup B)$ holds, as follows.
For sake of contradiction suppose $\exists z \in (A\cup B)$, such that $P(z)$ is false. It follows that $z \in A$, $z \in B$ or $z\in (A \cap B)$. In each of these cases $P(z)$ is true by assumption, so we arrive at a contradiction. Therefore $P(A\cup B)$ must be true. This completes my argument.
If anyone would be kind enough to verify my argument or point out any error(s), I would be grateful.