If $A_i$ are subsets of a metric space and $B_n=\cup^{n}_{i=1}A_i$, prove that $\bar{B_n}=\cup^{n}_{i=1}\bar{A_i}$ for $n=1,2,3,...$ For my course in real analysis I'm working on the problem: 
Let $A_1,A_2,A_3,...$ be subsets of a metric space and $B_n=\cup^{n}_{i=1}A_i$, prove that $$\overline{B_n}=\cup^{n}_{i=1}\overline{A_i},\quad n=1,2,3,...$$
My attempt: (1) Let $x\in\overline{A_i}$ a limit point of $A_i$, then for any neighbourhood $N$ of $x$, $A_i\cap N\neq \emptyset$ so $N\cap B_n$ is not empty. Therefore $x\in B_n'$ and $x\in \overline{B}_n$.
(2) Let $x\in B_n'$, then for any neighbourhood $N$ of $x$, $N\cap(A_1\cup ... \cup A_n)\neq\emptyset$. So for any neighbourhood, $N\cap A_i\neq\emptyset$ for some $A_i$. Therefore $x$ is a limit point of this $A_i$.
Now I talked about the proof with one of the teachers why said that the second argument was invalid; but I didn't manage to understand why that would be the case. So I'm hoping someone here can explain the error better.
 A: The problem lies in the detail here:
To show that $x$ is a limit point of $A_j$, you have to show that $N \cap A_j \neq \emptyset$ for every $N$. What you have shown, however, is that for every $N$ there exists an $i$ such that $N \cap A_i \neq \emptyset$. Do you see the problem? You can't show that $j = i$ holds for all $N$, the $i$ might differ between $N$.
To repair your proof, you might want to show/use that a finite union of closed sets is closed. Then use that the closure is the minimal closed set, such that...
A: Proceed by induction. The equality is trivially true for $n= 1.$ For $n=2$ we have:
$\overline{B_2} = \overline{A_{1}\cup A_{2}}$
$= (A_{1}\cup A_{2})\cup (A_{1}\cup A_{2})'$
$= (A_{1}\cup A_{2})\cup (A_{1}'\cup A_{2}')$
$=  (A_{1}\cup A_{1}')\cup (A_{2}\cup A_{2}')$
$=\overline{A_1}\cup\overline{A_2}$.
Let the statement be true for some $n=k$.
Now consider $B_{k+1} = \displaystyle \cup_{i=1}^{k+1}A_i$
$\Rightarrow \overline{B_{k+1}} = \overline{\displaystyle \cup_{i=1}^{k+1}A_i}$
$=\displaystyle\overline{\cup_{i=1}^{k}A_i\cup A _{k+1}}$
$=\displaystyle \overline{\cup_{i=1}^{k}A_i}\cup\overline{A_{k+1}}$
$= \overline{B_{k}}\cup\overline{A_{k+1}}$
$= \displaystyle\cup_{i=1}^{k}\overline{A_i}\cup\overline{A_{k+1}}$
$=\cup_{i=1}^{k+1}\overline{A_{i}}$
Note: for the case of $n=2$ we used the fact that $(A_{1}\cup A_{2})'= (A_{1}'\cup A_{2}')$ which can be easily proved as below:
Consider $x \in (A_{1}\cup A_{2})'$.
$\iff \forall \, \varepsilon >0 \, \, N_{\varepsilon}(x) \cap ((A_{1}\cup A_{2})\setminus \{x\}) \neq \emptyset$
$\iff \forall \, \varepsilon >0 \, \, N_{\varepsilon}(x) \cap (A_{1} \setminus \{x\})\cup(N_{\varepsilon}(x) \cap (A_{2} \setminus \{x\})\neq \emptyset$
$\iff x$ is a limit point of $A_{1}$ or $A_{2}$
$\iff x \in A_{1}'\cup A_{2}'$
