Closure of Union contains Union of Closures I'm teaching my self topology using a book I found. This is the second part of a 4 part question. First part is here.
I'm trying to prove the following problem from a book I found:

Let $X$ be a topological space and let $\mathscr{A}$ be a collection of subset of $X$. Prove  $\overline{ \bigcup \limits_{A \in \mathscr{A}} A}\supseteq \bigcup \limits_{A \in \mathscr{A}} \overline{A}$

*where the over line indicates closure.
First I'm going to show an example the them being equal, and then show an example of a proper superset. I know the examples are not needed for the proof, but I find them insightful, and these posts are like a notebook that I can refer back to later. Then I will attempt to prove this.

(I) Example of $\overline{ \bigcup \limits_{A \in \mathscr{A}} A} = \bigcup \limits_{A \in \mathscr{A}} \overline{A}$

I find that with "most" collections of subsets, the two sides come up equal. Let $X$ be $\mathbb{R}$ with the standard Euclidean metric. Also let, $\mathscr{A}=\{(1,6),(3,10)\}$. So then $ \bigcup \limits_{A \in \mathscr{A}} A= (1,10)$, and  $\overline{ \bigcup \limits_{A \in \mathscr{A}} A} =[1,10]$. Next the right side would be: $\bigcup \limits_{A \in \mathscr{A}} \overline{A}= [1,6] \cup [3,10]=[1,10]$. So it is possible for the two sides to be equal.

(II) Example of $\overline{ \bigcup \limits_{A \in \mathscr{A}} A} \supset \bigcup \limits_{A \in \mathscr{A}} \overline{A}$

The only way I know of involves a collection of subsets that "expands" in a way that converges - ie  $\bigcup_{n=1}^{\infty} \overline {B_{\frac{R \cdot n}{n+1}}}(x)=B_{R}(x)$. Let $X$ be $\mathbb{R}$ with the standard Euclidean metric. Also let, $\mathscr{A}=(1,\frac{10n}{n+1}), n\in\mathbb N$. So then $ \bigcup \limits_{A \in \mathscr{A}} A= (1,10)$, and  $\overline{ \bigcup \limits_{A \in \mathscr{A}} A} =[1,10]$. 
Next the right side would be: $\bigcup \limits_{A \in \mathscr{A}} \overline{A}= \lim \limits_{n \rightarrow \infty} [1, \frac{10n}{n+1}]= [1,10)$ . And $[1,10] \supset [1,10)$. So it is possible for Closure of Unions to contain Union of Closures.

Proof: $\overline{ \bigcup \limits_{A \in \mathscr{A}} A}\supseteq \bigcup \limits_{A \in \mathscr{A}} \overline{A}$

(II) illustrates that the infinite union of closed sets can make an open (or half open) set. So:
* $\bigcup \limits_{A \in \mathscr{A}} \overline{A}$ can be (half)open.
 * $\overline{ \bigcup \limits_{A \in \mathscr{A}} A}$ must be closed.
Since we're dealing with the same collection of subsets, then the difference will be at the boundary points. And thus, the closed set will contain the (half)open set.
QED
The part of the proof i'm not sure about is at the end where I say, "the difference will be at the boundary points." In (II) that is the case, but will that always be the case when dealing with the most general topological spaces?
 A: ‘[T]he difference will be at the boundary points’ doesn’t really say anything: the boundary points of what? Some particular $A\in\mathscr{A}$? $\bigcup\mathscr{A}$? $\bigcup_{A\in\mathscr{A}}\operatorname{cl}A$?
Just show directly that 
$$\bigcup_{A\in\mathscr{A}}\operatorname{cl}A\subseteq\operatorname{cl}\bigcup\mathscr{A}$$
by showing that if $x\in\bigcup_{A\in\mathscr{A}}\operatorname{cl}A$, then $x\in\operatorname{cl}\bigcup\mathscr{A}$.
Suppose that $x\in\bigcup_{A\in\mathscr{A}}\operatorname{cl}A$; then there is some $A_x\in\mathscr{A}$ such that $x\in\operatorname{cl}A_x$. Let $U$ be any open nbhd of $x$; then
$$U\cap\bigcup\mathscr{A}=\bigcup_{A\in\mathscr{A}}(U\cap A)\supseteq U\cap A_x\ne\varnothing\;.$$
Thus, every open nbhd of $x$ intersects $\bigcup\mathscr{A}$, and therefore $x\in\operatorname{cl}\bigcup\mathscr{A}$, as desired.
A: (1). Observe that $A\subset Y\implies \bar A\subset \bar Y.$
(2). Observe that $(\;\forall B\in \mathbb B\;(B\subset C)\;)\implies \cup_{B\in \mathbb B}B\subset C.$
(3). We have $\forall A\in \mathbb A \;(A\subset \cup_{A\in \mathbb A}A).$ So by (1) with $Y=\cup_{A\in \mathbb A}A ,$ we have $$\forall A\in \mathbb A\;(\bar A\subset \overline {\cup_{\mathbb A}A}).$$
(4). Applying (2) to the result of (3), with $B=\bar A$ and $\mathbb B=\{\bar A :A\in \mathbb A\}$  and $C=\overline {\cup_{\mathbb A}A},$ we have the desired result.
(5). A typical example of strict inequality is  $\mathbb A=\{\{a\}:a\in A\}$ where $A$ is  a non-closed subset of a $T_1$ space, that is, a space in which $1$-element subsets are closed. Then $\overline {\cup \mathbb A}=\bar A\ne A=\cup_{A\in  \mathbb A}\bar A.$
(5). Remark. Another useful general property is that Cl$( \cup \mathbb A) =$Cl$( \cup \{\bar A: A\in \mathbb A\} ).$
