TITLE: Countable ordinals are embeddable in the rationals $\Bbb Q$ -- proofs and their use of AC
QUESTION [10 upvotes]: Yesterday, Asaf Karagila's answer to my question sparked an extensive discussion on ways of proving that all countable ordinals are embeddable in $\Bbb Q$, and whether particular solutions to this use the Axiom of Choice.
Since the arguments used to prove this can be shady on their use of Choice when formulated tersely, fully fleshed out answers are appreciated.
To be concrete, this question asks for proofs of the following fact:

All countable ordinals $\alpha < \omega_1$ are order-embeddable in the rationals $\Bbb Q$.

Especially answers avoiding the Axiom of Choice are appreciated.

Of relevance and related interest is #123969 (but it's not a duplicate, since that question asks for explicit embeddings).

REPLY [9 votes]: Finite linear orders are no problem, so let $\langle X,\preceq\rangle$ be any countably infinite linear order, and fix an enumeration $X=\{x_k:k\in\omega\}$. Also fix an enumeration of the rationals, $\Bbb Q=\{p_k:k\in\omega\}$. Define $f:X\to\Bbb Q$ as follows.
Let $f(x_0)=p_0$. Suppose that $0<n\in\omega$, and $f(x_k)$ has been defined for each $k<n$. Let $L_n=\{k<n:x_k\prec x_n\}$ and $R_n=\{k<n:x_n\prec x_k\}$. 

If $L_n\ne\varnothing\ne R_n$, let $\ell(n)=\max\{x_k:k\in L_n\}$ and $r(n)=\min\{x_k:k\in R_n\}$. Then $x_{\ell(n)}\prec x_{r(n)}$, so $f(x_{\ell(n)})<f(x_{r(n)})$, and we can set $f(x_n)=p_m$, where $$m=\min\{k\in\omega:f(x_{\ell(n)})<p_k<f(x_{r(n)})\}\;.$$
If $L_n=\varnothing\ne R_n$, let $r(n)=\min\{x_k:k\in R_n\}$, and set $f(x_n)=p_m$, where $$m=\min\{k\in\omega:p_k<f(x_{r(n)})\}\;.$$
And if $L_n\ne\varnothing=R_n$, let $\ell(n)=\max\{x_k:k\in L_n\}$, and set $f(x_n)=p_m$, where $$m=\min\{k\in\omega:f(x_{\ell(n)})<p_k\}\;.$$

Given the enumerations, everything is explicitly defined.