How can a set be uncountable but well-ordered? By well-ordering theorem (from axiom of choice), any set can be well-ordered. Well-ordering says a set can be counted from bottom up, but how can such a set be uncountable?
 A: Consider the set $\mathbb{N}\cup\{\omega\}$, which is well-ordered by the order defined by $x\le y$ if $y=\omega$ or $x,y\in\mathbb{N}$ and $x\le y$.
This is a well-ordered set, but you cannot use the order on it to inductively enumerate its elements. Indeed, by taking the least element and continuing, you will only be able to get $\{1,2,3,\ldots\}=\mathbb{N}$.
I'm not claiming that this set is uncountable (because it clearly isn't). I only hope this gives some intuition on how an uncountable well-ordered set could possibly exist if one accepts the axiom of choice.
Edit: As pointed out by Hanul Jeon in the comments, the existence of an uncountable well-ordered set does not require the axiom of choice. (See here: A well-order on a uncountable set). However it is required to well-order ANY set (via the well-ordering theorem).
A: There's a wrong assumption there. Well ordering doesn't mean that “a set can be counted from bottom up”. It means that every non-empty subset has a least element.
A: Notice also that there are uncountable ordinals (see First uncountable ordinal; see also Simple example of uncountable ordinal).
Every ordinal is a well-ordered set (by the relation $\in$).
If there are uncountable ordinals, then clearly there are uncountable well-ordered sets.
