Let $A\subsetneq[0,1]$ be some countable set of real numbers. Since the rationals are dense in the reals and since all countable linear orders are embeddable into $\mathbb{Q}$, it seems to me that there has to be a map $f:A\rightarrow[0,1]_\mathbb{Q}$ with $[0,1]_\mathbb{Q}$ being the rational interval such that for any $a\in A$: $$a\leq a'\text{ iff }f(a)\leq f(a')$$
Is it possible to construct $f$ explicitly, without resorting to the axiom of choice?
$f$ does not have to be onto.
From my limited understanding of set theory, you can do this with induction without using any variant of the axiom of choice at all.
Since $A$ is countable, you are assuming there exists some enumeration of its elements $\{a_n\}_{n\in\mathbb{N}}$. Start with the empty map mapping the empty subset of $A$ to an empty subset of $\mathbb{Q}\cap [0,1]$. The induction step is assuming you already have a mapping of $a_1,...,a_k$ to some rationals in this interval $b_1, ..., b_k$ that preserves the linear order. To perform the induction step you need to choose a rational to send $a_{k+1}$ to. Ok. There's a unique pair $a_s, a_t$ such that $s, t \le k$, $a_s$ and $a_t$ are adjacent in the linear order and $a_{k+1}$ is between them. So Let $b_{k+1} = (b_s+b_t)/2$. You need to handle the case where $a_{k+1}$ is to the left, or right of all $\{a_1, ..., a_k\}$ and make sure you don't go beyond the interval $[0,1]$. I'll let you complete the detail, but the point is there is no use of the axiom of choice because you're giving an algorithm how to choose $b_{k+1}$.
To finish the proof, you take the union of all finite partial maps you constructed this way to get a complete mapping of $\{a_n\}_{n\in\mathbb{N}}$ to $\{b_n\}_{n\in\mathbb{N}}$.