Is a well-ordered set always countable? Suppose that $A$ is a well-ordered set. We can find $a_{0} = \min{A}$. Then, if $A \setminus \left\{a_{0}\right\} \neq \emptyset$, we can find $a_{1} = \min{\left(A \setminus \left\{a_{0}\right\}\right)}$, etc. Intuitively, this seems to be a counting process. Then my hypothesis is, every well-ordered set is countable. Is this a valid guess? In my understanding, if it is valid, then for finite well-ordered sets, we can use mathematical induction to prove it. Then what about infinite cases?
 A: To clarify exactly what's going on, your argument shows there is a countably infinite subset of any infinite well-ordered ordinal. The reason it doesn't show any well-ordered set is countable is that there's no reason your process should exhaust the set A. If we consider the set $\mathbb{N}\cup\{\infty\}$, where we add a point larger than any natural, we can check this is still a well-ordered set. But the process you describe will never "count" the point at infinity. Of course this set is still countable, but the idea is instead of adding 1 point we can add uncountably many (a priori it's not clear how to order these points, this is just supposed to show why we can get uncountable ordinals).
A: Every ordinal is well-ordered by $\in$. Ordinals are arbitrarily large. All the axiom of choice adds is that all sets biject with an ordinal, thereby being well-orderable. Even proper classes can be well-orderable; the class of ordinals is well-ordered by $\in$.
A: I'd like to point out why your intuitive argument is wrong. Your sequence starts with the smallest element of $A$, then moves to the next smallest element of $A$ - the successor of $a_0$. By induction (as per the Peano axioms), this process will eventually reach every natural number if we apply it to $\mathbb N$. By the way, it's typical to write $\omega = \mathbb N$ when referring to it as an ordinal. The thing is, your process being exhaustive would require that every element of $A$ other than the smallest one is a successor of some other element. I'll give you an example of a well ordered set where this does not hold.
As I said, we have $\omega = \mathbb N$. Let's define a new set, called $\omega + 1$, as $\omega \cup \{\infty\}$, where $\infty$ is just some symbol not in $\omega$. We'll keep the ordering on $\omega$ and assert that $\infty$ is strictly bigger than every element of $\omega$. This makes $\omega + 1$ into a well ordered set. Essentially, if a nonempty subset contains $\infty$ then its minimum is either $\infty$ itself or removing $\infty$ does not change the minimum. Now, if you tried to apply this process to $\omega + 1$, you would get $a_0 = 0$, $a_1 = 1$,... , $a_n = n$. However, you would never reach $\infty$ as there is no $n \in \omega$ whose successor is $\infty$.
I'd like to point you to the notion of transfinite induction, however. This is a method that allows you to make inductive arguments on any well ordered set, as you have tried to do. The difference is that you have to account for these $\infty$ esque elements which are not a successor of anything (called limit ordinals). If you can prove a property $P$ holds for all limit ordinals in $A$ and that $P(\alpha) \implies P(succ(\alpha))$ then $P$ holds for all $\alpha$ in $A$.
I'd also like to point out that the more standard notation is $\omega + 1 = \omega \cup \{\omega\}$. This definition of the successor ordinal allows for a very clean theory. Here, we have that $\omega \in \omega + 1$ is a limit ordinal, so doing induction on $\omega + 1$ requires treating this case specially.
