Use the fact $x \in A' \iff \forall r > 0 (B(x,r)\bigcap A \text{ is infinite })$ to show $(A \bigcup B)' \subseteq A' \bigcup B'$. Where $A'$ is the derived set. We note:
$$x \in A' \iff \forall r > 0 (B(x,r)\bigcap A \text{ is infinite })$$
Am I missing something here?
Suppose $x \in (A \bigcup B)'$. Then $\forall r > 0, B(x,r) \bigcap (A \bigcup B)$ contains infinite points of $A$ or $B$. This implies $B(x,r) \bigcap A$ is infinite or $B(x,r) \bigcap B$ is infinite. So $x \in A'$ or $a \in B'$. Thus $(A \bigcup B)' \subseteq A' \bigcup B'$.
 A: Your proof is wrong, because you need the same set ($A$ or $B$) for all $r > 0$. Now it could be that half of the time you'd have an infinite intersection of the ball with $A$, and another half of the time an infinite intersection with $B$, but not the same one for all $r$. In that case the proof would break down.
So to show the inclusion, we take any $x \in (A \cup B)'$. We have to show that $x \in A' \cup B'$. So suppose not: then $x \notin A'$, and $x \notin B'$.
By the property of infinite intersection, this means that there is some $r_1 > 0$ such that $B(x,r_1) \cap A$ is finite (as $ \notin A'$) and also some $r_2 > 0$ such that $B(x, r_2) \cap B$ is finite.
Now let $r = \min(r_1,r_2) > 0$. Now:
$$B(x, r) \cap (A \cup B) = (B(x,r) \cap A) \cup (B(x,r) \cap B) \subseteq (B(x,r_1) \cap A) \cup (B(x,r_2) \cap B)$$
and the right hand side is finite, as a union of two finite sets.
It follows, by the same property, that $x \notin (A \cup B)'$, which is a contradiction to how $x$ was chosen. So $x \in A' \cup B'$, as required.
