Let $A'$ denote the set of limit points of $A$ where $A$ is a subset of a metric space $X$
Is the claim in the title true? And if so, is the following proof correct (just the inclusion $\subset$)?
Let $x \in (A \cup B)'$. Then for every $B_r(x)$, there exists $y$ such that $x \ne y$ and $y \in B_r(x) \cap (A \cup B)$. Hence $y \in A$ or $y \in A$ or $y \in B$ or both. If $y \in A$, then $x \in A'$ and if $y \in B$, then $x \in B'$. Thus $x \in A' \cup B'$. So $(A \cup B)' \subset A' \cup B'$