Proving that the union of the limit points of sets is equal to the limit points of the union of the sets My instructor gave us an exercise to do, to show that this equality holds:
$$\bigcup_{k=1}^m A_k'=\left(\bigcup_{k=1}^m A_k\right)^{\!\prime}.$$
My thoughts on approaching this question is by contradiction, suppose that there is an element that satisfies the left side of the equality and not the right side. Then there is a deleted ball with radius $r$ such that $B(x,r)$ intersect the left side of the equality is non empty while there is a deleted ball with radius $r$ such that $B(x,r)$ intersect the right side of the equality is empty. Is this the right approach in proving this? This is as far as I can go.
Thanks for the help in advance.

A slightly different way to prove
$\left(\bigcup_k A_k\right)'\subseteq \bigcup_k A_k'$
is as followed.
Say $p\in\left(\bigcup_k A_k'\right)^c$, i.e. $p$ is not a limit point of any of $A_k$, there exists a neighborhood $N(p)$ (a ball of radius $r$ centered at $p$, if you will) that does not intersect any of $A_k$.
This means $N(p)$ does not intersect $\bigcup_k A_k$ either, so therefore, $p$ cannot be a limit point of $\bigcup_k A_k$, i.e. $p\in\left(\left(\bigcup_k A_k\right)'\right)^c$.
As a result,
$\left(\bigcup_k A_k'\right)^c \subseteq \left(\left(\bigcup_k A_k\right)'\right)^c,$
so
$\left(\bigcup_k A_k\right)' \subseteq \bigcup_k A_k'$
