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.