Suppose I) X = $\mathbb{N}$ II) $X_{\alpha}$ = {x $\ge$ $\alpha$ | x $\in$ $\mathbb{N}$} Hence, $\cap_{\alpha}^N$ $X_{\alpha}$ = [N, $\infty$).By II, all $X_{\alpha}$ are well ordered sets with a minimum of $\alpha$. Hence the union of all $X_{\alpha}$ is the well ordered set of natural numbers, which shows all subsets of $\mathbb{N}$ are well ordered sets.
Is this a correct way to show all subsets of a well ordered set are well ordered sets?
On
You seem to argue that an infinite union of well-ordered sets has to be well-ordered. But this is not true.
Also, what is your argument that $X_a$ is really well-ordered? Is it because it is just a translated copy of $\Bbb N$? That would make the argument circular! Or is it because the $X_a$ themselves have a minimal element? Well-orderedness is about all non-empty subsets having a minimal element!
Your proof is not clear, but you can use this idea.
A set $X$ together with ordering $\leq$ is called a well-ordered set if every non-empty subset of $X$ has the smallest element. Let $S$ be a subset of $X$, since then for every subset $A$ of $S$ it is a subset of $X$, hence it has a least element. Therefore $S$ is also a well-ordered set.