The rationals in an interval are countably infinite while the interval is uncountable

${\mathbb{Q}}_{\mathbb{R}}$ is countably infinite. Since $\epsilon$ is injective, it is a bijection of $\mathbb{Q}$ onto its image ${\mathbb{Q}}_{\mathbb{R}}$; by [F1] there is a bijection $\mathbb{Q}\to \mathbb{N}$, and composing its inverse with $\epsilon$ exhibits a bijection $e:\mathbb{N}\to {\mathbb{Q}}_{\mathbb{R}}$. In particular ${\mathbb{Q}}_{\mathbb{R}}$ is at most countable.

$S_{a,b}$ is at most countable, nonempty, and dense in $I_{a,b}$. It is a subset of ${\mathbb{Q}}_{\mathbb{R}}$, so [F3] applies; it is nonempty and dense by [F2].

The step function $F$ is well defined on $I_{a,b}$. Fix the bijection $e:\mathbb{N}\to {\mathbb{Q}}_{\mathbb{R}}$ of step 1.1. For $x\in I_{a,b}$ the set $\{k\in \mathbb{N}:x<e(k)<b\}$ is nonempty by [F2] applied to $x<b$, and it is a subset of $\mathbb{N}$, so it has a least element $k(x)$ by [F4]; put $F(x):=e(k(x))$, a rational with $x<F(x)<b$. Likewise $\{k:a<e(k)<b\}$ is nonempty by [F2] applied to $a<b$, and its least element $k_0$ gives the seed $r_0:=e(k_0)\in S_{a,b}$.

The recursion $r(0):=r_0$, $r(\sigma (n)):=F(r(n))$ produces a strictly increasing injection $r:\mathbb{N}\to S_{a,b}$. Since $F$ maps $I_{a,b}$ into itself — $x<F(x)<b$ and $a<x$ give $a<F(x)$ — the recursion theorem applies with $X=I_{a,b}$ by [F4], and $r(n)\in I_{a,b}$ for all $n$. Moreover $r(n)<r(\sigma (n))$ for every $n$ by the defining property of $F$, Inductively, $m<n$ implies $r(m)<r(n)$: the case $n=0$ is vacuous, and $m<\sigma (n)$ gives $m\le n$, so either $m=n$ and the adjacent inequality applies, or $m<n$ and transitivity combines the induction hypothesis with $r(n)<r(\sigma (n))$. Distinct indices are comparable, hence $r$ is injective. Finally every value $r(n)$ lies in ${\mathbb{Q}}_{\mathbb{R}}\cap I_{a,b}=S_{a,b}$.

Part (ii): $C:=I_{a,b}\setminus {\mathbb{Q}}_{\mathbb{R}}$ is uncountable. Suppose $C$ were at most countable. If $C=\mathrm{\varnothing }$ then $I_{a,b}=S_{a,b}$ is at most countable by step 2.1, contradicting [F6]. Otherwise $C\mathrm{\ne }\mathrm{\varnothing }$, and then [F3] applies to $S_{a,b}$ and to $C$, which are both nonempty at most countable — $S_{a,b}$ by step 2.1 — giving surjections $f,g:\mathbb{N}\to S_{a,b},C$. Define $t:\mathbb{N}\times \mathbb{N}\to I_{a,b}$ by $t(0,k):=f(k)$ and $t(\sigma (i),k):=g(k)$; every element of $I_{a,b}$ lies in $S_{a,b}$ or in $C$ by [F7], so $t$ is onto. Composing $t$ with the bijection $\mathbb{N}\to \mathbb{N}\times \mathbb{N}$ of [F7] gives a surjection $\mathbb{N}\to I_{a,b}$, so $I_{a,b}$ is at most countable by [F3], contradicting [F6]. Hence $C$ is uncountable.

Therefore $\mathbb{N}⪯S_{a,b}$ via the injection $r$, and [F3] supplies a surjection $h:\mathbb{N}\to S_{a,b}$ since $S_{a,b}$ is nonempty and at most countable by step 2.1. The rule $x\mapsto \min \{k:h(k)=x\}$ is an injection $S_{a,b}\to \mathbb{N}$ by [F4]: equal least indices have the same $h$-value. Schröder-Bernstein therefore gives $S_{a,b}\approx \mathbb{N}$; this is (i).

Collecting: $S_{a,b}\approx \mathbb{N}$ by step 4.1, so the rationals of the interval are countably infinite; $S_{a,b}$ is dense in $I_{a,b}$ by step 2.1; and $I_{a,b}\setminus {\mathbb{Q}}_{\mathbb{R}}$ is uncountable by step 3.2 while $I_{a,b}$ is uncountable by [F6]. The increasing injection $r$ is defined by least elements of nonempty sets of naturals and a single recursion, and the two surjections in the interleaving are given by the criterion, so no choice principle is used.