Show that the union of complete sets is not necessarily complete $\textbf{Proof: }$ Let $X=\mathbb{R}$ and consider $A_n=[\frac{1}{n},1],n\in\mathbb{N}$. We have that  $A_\delta$ closed interval in $\mathbb{R}$ are complete. 
Thus $A_n$ is complete in $\mathbb{R},\forall n\in\mathbb{N}$.
Then $\displaystyle\bigcup_{n\in\mathbb{N}}{A_n}=\left(0,\frac{1}{n}\right]$.
$A_\delta\{\frac{1}{n}\}$ is a Cauchy sequence in $\displaystyle\bigcup_{n\in\mathbb{N}}{A_n}$ however it does not have a limit point in $\displaystyle\bigcup_{n\in\mathbb{N}}{A_n}$. 
Thus $\displaystyle\bigcup_{n\in\mathbb{N}}{A_n}$ is not complete. 
I think I got the gist of it, maybe I'm missing some key factors or there could be a better way of proving this. Any advice is appreciated. Thanks. 
 A: A counterexample is a perfectly valid way of showing that something is not always the case, and you have a perfectly good counterexample.
As for your proof in particular, I don't know what $A_\delta$ is. Actually, if you take the proof that you have, and just erase the two $A_\delta$ that you've written, then it reads just fine, apart from a few grammatical hiccups that have nothing to do with math or logic.
Edit: Apparently I missed that you'd written $\bigcup A_n=(0,1/n]$. That should, of course, be $(0,1]$.
A: What you did is fine, except that $\bigcup_{n\in\mathbb N}A_n=(0,1]$.
A simpler example would be to take $A_n=\left\{\frac1n\right\}$. Then each $A_n$ is complete, but$$\bigcup_{n\in\mathbb N}A_n=\left\{\frac1n\,\middle|\,n\in\mathbb N\right\},$$which is not.
A: This proof seems to me unnecessarily complicated. You asked for "a better way of proving this"; here is one.
Let $X$ be any non-complete set.
Let $A_x=\{x\}$ for each element $x$ in $X$.
Then each $A_x$, being finite, is trivially complete, but their union is $X$ which is incomplete.
(Daniel Wainfleet's example above is a special case of this.)
This is so much simpler that I wonder if you haven't misunderstood the question. Are you sure it wasn't asking you to prove that a finite union of complete sets must be complete?
