If $\lt$ is a well-ordering of $S$, then we may define the lexical ordering on the power set $P(S)$, by which $A\lt_{lex} B$, if and only if the $\lt$-least element of the symmetric difference $A\triangle B$ is in $B$ and not in $A$. That is, we look to the first place where the sets differ, and then put the set without this element before the set with this element. This is just like the order in a dictionary, hence the name, since two words in a dictionary are put in order by comparing the first letter on which they differ.
To see that this is a total order, we check first that it is transitive. If $A\lt_{lex}B\lt_{lex} C$, then the first difference between $A$ and $C$ must be in $C$, since either this occurs before the first difference between $A$ and $B$, in which case it is in $C$ and not in $B$ and hence in $C$ and not in $A$, or else it occurs at or after the first difference between $A$ and $B$, in which case it occurs exactly at this difference, which is in $C$ and not in $A$. The order it linear since any two sets do indeed have a least difference, since $\lt$ is a well-order.
You have to distinguish between two things:
- A set can be well-ordered.
- A set with a natural linear ordering is well-ordered with that natural ordering.
The axiom of choice implies that every set can be well-ordered. But of course not every set which has a natural linear order is well-ordered. You don't need to venture to the real numbers. Both $\Bbb Z$ and $\Bbb Q$ have a natural linear order which is most definitely not well-ordered.
The point is that given a set of real numbers, if it is well-ordered in the usual ordering of the real numbers, then it is countable. We can prove this by choosing a canonical rational point from each interval between a point and its successor. This choice of rational numbers is not using the axiom of choice, since we can always choose a canonical rational number from an interval (e.g., represent each rational as $\frac pq$ where $p,q\in\Bbb Z$, $p>0$ and $\gcd(p,q)=1$; then consider the rational with the smallest $p$ possible in the interval, and find the one with the denominator closest to $0$ (the positive if both options exist) that match this numerator).
And the main confusion people have with the well-ordering theorem, is that if $X$ is a set which has a natural linear ordering, then there is absolutely no reason for any well-ordering to agree with the natural linear ordering. Much like how we can define well-orderings of $\Bbb N$ which disagree with the usual ordering, or how we can define a well-ordering of the rational numbers which certainly disagrees with their natural order.
(What is true, as you mention, is that we cannot specifically define a set of real numbers in the language of set theory such that $\sf ZF$ proves that this set is both uncountable and can be well-ordered. Namely, it is consistent with the failure of the axiom of choice that only countable sets of real numbers can be well-ordered.)
Best Answer
There can't be. If $S\subseteq \Bbb R$ is well-ordered by the usual ordering, for every element $s_{\alpha}\in S$ that has an immediate successor $s_{\alpha+1}\in S$ (every element of $S$ except the greatest element if there is one), the set of rationals $Q_{\alpha}$ between the element and its successor is nonempty: $(s_{\alpha}, s_{\alpha+1}) \cap \Bbb Q \ne \emptyset$, and the $Q_{\alpha}$ are disjoint. If $S$ were uncountable, then $\bigcup_{\alpha < length(S)} Q_{\alpha}$ would also be uncountable — impossible, as it's a subset of $\Bbb Q$.