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I have a proof, not sure if it is correct.

Consider a limit point, $p$, of $( A \cup B)'$. Assume via contradiction that $p$ is a limit point of neither $A$ nor $B$. Then there exists two neighborhoods, $N_1$ and $N_2$ such that $p\in N_1$ and $p \in N_2$ and $N_1 \cap A \neq \emptyset$ and $N_2 \cap B \neq \emptyset$ but $N_1 \cap B = \emptyset$ and $N_2 \cap A = \emptyset$. Then take the intersection of $N_1$ and $N_2$. Then, $N_1\cap N_2$ is nonempty (since $p$ is in the intersection) and $(N_1\cap N_2)\cap A=\emptyset$ and $(N_1\cap N_2)\cap B=\emptyset$ Also, $(N_1\cap N_2)$ is open since it is the finite intersection of open sets. Thus we have found an open set that contains $p$ whose intersection with $(A\cup B)$ is empty this means $p$ is not a limit point of $(A\cup B)$ -a contradiction.

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This is a good proof. – Bombyx mori Jan 23 '13 at 5:41
It might be a good idea to state that the notation $C'$ means the derived set of $C$. I first took it to mean the complement of $C$, and thought the whole problem was crazy. – Gerry Myerson Jan 23 '13 at 5:43
You could prove this directly in a somewhat simpler way. Suppose for simplicity that the topology is first countable, so that limits are always sequential limits. Then given an infinite convergent sequence $\{x_n\} \subset A \cup B$ converging to a point $x \in (A \cup B)'$, infinitely many of the points belong to one of $A$ or $B$. Suppose this is true for $A$ for instance. Then the subsequence $\{x_n\} \cap A$ converges to $x$ and so $x \in A'$. Your proof is better, however, as you don't need to appeal to any properties of the topology, or to the notion of nets. – A Blumenthal Jan 23 '13 at 5:51
As a minor comment, note that your proof is really by contrapositive as opposed to contradiction. – arjafi Jan 23 '13 at 5:56

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