Show that $\bigcup _{\alpha \in J}\operatorname{cl} A_\alpha \subseteq\operatorname{cl}\bigcup _{\alpha \in J} {A_\alpha} $ $\newcommand{\cl}{\operatorname{cl}}$Let $\{A_\alpha\}_{\alpha\in J}$ be a collection of subsets of a topological space X.
Show that $\bigcup _{\alpha \in J}\cl A_\alpha\subseteq\cl\bigcup _{\alpha \in J} {A_\alpha} $ 
Give an example where equality fails. 
Let $x\in \bigcup_{\alpha \in J}\cl A_\alpha $ 
$\Rightarrow$ $x\in\cl A_\alpha $  for all $\alpha \in J$
Now what?
 A: Pick one $\beta\in J$ and observe that:
$A_{\beta}\subset\cup_{\alpha\in J}A_{\alpha}\subset\text{cl}\left(\cup_{\alpha\in J}A_{\alpha}\right)$.
The last set is closed so from this we are allowed to conclude that:
$\text{cl}\left(A_{\beta}\right)\subset\text{cl}\left(\cup_{\alpha\in J}A_{\alpha}\right)$.
This is true for every $\beta\in J$ so: 
$\cup_{\beta\in J}\text{cl}\left(A_{\beta}\right)\subset\text{cl}\left(\cup_{\alpha\in J}A_{\alpha}\right)$.
In $\mathbb{R}$ with its common topology you have $\text{cl}\left(0,1\right)=\left[0,1\right].$
Also you have $\left(0,1\right)=\cup_{x\in\left(0,1\right)}\left\{ x\right\} $
where $\text{cl}\left\{ x\right\} =\left\{ x\right\} $. So...
A: $\newcommand{\cl}{\operatorname{cl}}$HINT: The easiest way to show that $\bigcup_{\alpha\in J}\cl A_\alpha\subseteq\cl\bigcup_{\alpha\in J}A_\alpha$ is to show that if $x\notin\cl\bigcup_{\alpha\in J}A_\alpha$, then $x\notin\bigcup_{\alpha\in J}\cl A_\alpha$. If $x\notin\cl\bigcup_{\alpha\in J}A_\alpha$, then there is an open nbhd $U$ of $x$ such that $U\cap\bigcup_{\alpha\in J}A_\alpha=\varnothing$, and therefore ... ?
For an example showing that equality need not hold, note that if $A$ is any set, $A=\bigcup_{x\in A}\{x\}$.
