TITLE: Is a well-ordered subset of $(\mathbb{R},<)$ countable?
QUESTION [2 upvotes]: Suppose $X \subseteq \mathbb{R}$ is well-ordered by $<$, the normal ordering on $\mathbb{R}$. Is it then countable? If so, how to prove this. I've tried to set up a bijection as follows: 
$$
\phi : \mathbb{N} \to X : n \mapsto \ <\textrm{-least element of } X \backslash \phi(\mathbb{N} \upharpoonright n)
$$
I.e, $0$ maps to the least element of $X$, 1 maps to the least element of $X \backslash \{x_1\}$, $2$ maps to the least element of $X \backslash \{x_1,x_2\}$ etc. This is injective by construction, but I'm not sure if it is surjective.

REPLY [1 votes]: Assume W is an uncountable, well ordered subset of R.
For each x in W, Let x' be the successor of x in the well order.
{ (x,x') : x in W } is a uncountable collection of pairwise disjoint open intervals,
That is impossible because in each of those intervals there is a different rational and there are only countablely many rationals but uncountable many intervals.
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