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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 $$(A \cup B)' \subset A' \cup B'$$

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Your proof is not very valid. If you want to show $x \in A' \cup B'$, you need to show that for a given $r$, you can always pick $y \in A \cap B_r(x)$, or always pick $y \in B \cap B_r(x)$. You cannot alternate between the two sets as $r$ varies. – Tunococ Sep 10 '12 at 21:14
It still needs a little work. For each $r>0$ you get a $y_r$ that’s in either $A$ or $B$. If they’re all in $A$, then certainly $x\in A'$, and similarly if they’re all in $B$, but some $y_r$’s may be in $A$ while others are in $B$. You have to argue a bit more to cover this possibility. It can be done with what you have, but it’s easier to suppose that $x\in(A\cup B)'$ and $x\notin A'$ and prove that this forces $x$ to be in $B\,'$. – Brian M. Scott Sep 10 '12 at 21:14

Think about sequences. If $a\in A'$, you can find a sequence in $A$ that converges to $a$. The same sequence, seen now as a sequence in $A \cup B$ says that $a\in (A\cup B)'$.

Alternatively, if $d\in(A\cup B)'$, you can find a sequence $(d_i)$ in $A\cup B$ that converges to $d$. So any subsequence must also converge to $d$. Then, take either $(d_i)\cap A$ or $(d_i)\cap B$ as the subsequence which converges to $d$, showing that $d$ is also in either $A'$ or $B'$ (this depends on 'how many terms' in $(d_i)$ are in $A$ or $B$), so that $d\in A'\cup B'$.

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