If $x$ is the limit point of a finite union then is $x$ a limit point of some set in the union? Background
This question arose from the following problem: 
If $B_n=\bigcup_{i=1}^{n}A_i$ then prove that $\overline{B_n}=\bigcup_{i=1}^n \overline{A}_i$ for $n=1,2,3...$. By $\overline{X}$ we mean the closure of $X$, by $X'$ we mean the set of limit points of $X$ and by $N_r(x)$ we mean the neighborhood of $x$ of radius $r$.
Question and Attempt
I have done this problem before by induction but today I wanted to try doing it directly. 
Here is what I did:
We have to show that $$\left( \bigcup_{i=1}^n A_i \right)\bigcup\left( \bigcup_{i=1}^n A_i \right)'= \bigcup_{i=1}^n \left(A_i \cup A'_i \right) .$$ If $x \in \left( \bigcup_{i=1}^n A_i \cup A'_i \right)$ Then it is not too hard to show $x \in \left( \bigcup_{i=1}^n A_i \right)\bigcup\left( \bigcup_{i=1}^n A_i \right)'$. On the other hand though, suppose $x \in \left( \bigcup_{i=1}^n A_i \right)\bigcup\left( \bigcup_{i=1}^n A_i \right)'$ and consider the case when $x \in \left( \bigcup_{i=1}^n A_i \right)'.$ Then for all $r$, $N_r(x) \bigcap \left(\bigcup_{i=1}^n A_i \right)$ is not empty (in fact it contains infinitely many points since $x$ is a limit point). So for any $r$, this intersection contains infinitely many points of some $A_j$  ($1 \leq j \leq n$). 
How do we then conclude that $x$ is a limit point of some $A_j$, bearing in mind that as $r$ is changed, $N_r(x)$ may intersect a different set $A_k$?
 A: Yes. 
Suppose not. Let $x$ be a limit point of $ A\cup B$, but not a limit point of $A$ nor $B$. Then exist $r_1,r_2>0$, s.t. $B_{r_1}(x)\setminus \{x\}\cap A =\emptyset$ and $B_{r_2}(x)\setminus \{x\}\cap B =\emptyset$. Then take $r=\min \{r_1,r_2\}>0$, we get  $B_{r}(x)\setminus \{x\}\cap (A\cup B) =\emptyset$. Contradiction.
This does not generalize to infinite union. For example, $0$ is a limit point of $\bigcup [\frac{1}{n},1]$, but it's not a limit point for any of them.
You can also prove the theorem you want by using closure of $E$ is the smallest closed set containing $E$. Hence $\overline{B_n} \subset \bigcup_{i=1}^n \overline{A}_i$ since finite union of closed set is closed. You can easily get the other side inclusion.
A: Let it be that $x\notin\bigcup_{i=1}^{n}\overline{A}_{i}$. 
Then for each
$j\in\left\{ 1,\dots,n\right\} $ there is some $r_{j}>0$ such that
$N_{r_{j}}\left(x\right)\cap A_{j}=\emptyset$. 
For $r=\min\left\{ r_{1},\dots,r_{n}\right\}>0$
we then find: $N_{r}\left(x\right)\cap\left(\bigcup_{i=1}^{n}A_{i}\right)=\emptyset$.
This allows the conclusion that $x\notin\overline{B}_{n}$.
Proved is now: $x\in\overline{B}_{n}\Rightarrow x\in\bigcup_{i=1}^{n}\overline{A}_{i}$.
