$\textbf{Proof: }$ 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 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 $(0,1]$.