Is there any well-ordered uncountable set of real numbers under the original ordering? I know that the usual ordering of $\mathbb R$ is not a well-ordering but is there an uncountable $S\subset \mathbb R$ such that S is well-ordered by $<_\mathbb R$? 
Intuitively I'd say there is no such set but intuitively I'd also say there is no well-ordered uncountable set at all, which is obviously wrong. I still struggle to grasp the idea of an uncountable, well-ordered set.

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 $\Bbb Q$.
