Existence of a particular well-ordering of [0,1]

Question

How do you show, assuming the Axiom of Choice and the Continuum Hypothesis, that there exists a well-ordering on $[0,1]$ such that for all $x$, there are only countably many $y$ such that $y \leq x$?

Answer 1

If CH holds and AC both hold, then $[0,1]$ (which is bijectable with $\mathbb{R}$, hence with $2^{\aleph_0}$) is bijectable with $\omega_1$, the first uncountable ordinal. Let $f\colon [0,1]\to\omega_1$ be a bijection, and define the order with $x\leq y\iff f(x)\preceq f(y)$ (the right hand side is the usual ordering of ordinals).

Since $\omega_1$ is the first uncountable ordinal, every element of $\omega_1$ has only countably many elements strictly smaller than it, so for every $\alpha\in\omega_1$, $\{a\in\omega_1\mid a\preceq\alpha\}$ is countable. Thus, for any $x\in [0,1]$, only countably many reals can be strictly smaller than $x$ in this ordering.