Does a special property hold if the Archimedean property for reals doesn't hold?

Question

Suppose $\mathbb{R}^e=A \cup B$ in which $A \cap B=\varnothing$ and there exist real numbers $a_0$ and $b_0$ such that $a_0 \in A$ and $b_0 \in B$. My question is, can we construct $a \in A$ and $b \in B$ such that there exists a natural number $n$ such that $|a-b|<n$?

The restrictions is that I don't want to use the law of the excluded middle and the axiom of countable choice. My motivation is that $\mathbb{R}^e$, classical real numbers described by Troelstra and van Dalen in the book constructivism in mathematics, is not Archimedean. It is just weakly Archimedean that is:

$$ \forall x\in\mathbb{R}^e ~ \neg\neg\exists n\in \mathbb{N} ~ x<n. $$

What I already know about the problem is that, we can assume, without loss of generality, given any rational number $r$, $b_0+r \in B$. This is because either $b_0+r \in A$ or $b_0+r \in B$. If $b_0+r \in A$ we can solve the problem. So we can assume that $b_0+r \in B$.

Answer 1

We can assume, without loss of generality, that $-1 \in A$. Consider $|b_0|$. Either $|b_0| \in B$ (case 1) or $|b_0| \in A$ (case 2).

Case 1:

Put $$ X=\{x \in A: x < |b_0|\}. $$ According to the book Constructivism in mathematics by Troelstra and van Dalen, every inhabited subset of $\mathbb{R}^e$ which is weakly bounded above, has a supremum. So, since $-1 \in A$, $X$ is inhabited and since it is bounded above, it has a supremum. Put $s=sup(X)$. Either $s \in A$ or $s \in B$. If $s \in B$, since $s$ is the supremum we have $$ \exists x \in X (s-1<x). $$ Since $s$ is the supremum, $x \le s$. So we have $s-1<x \le s$. Therefore $|x-s|<1$ and we are done. If $s \in A$, consider $s+1$. Either $s+1 \in A$ or $s+1 \in B$. If $s+1 \in B$, since $|s-(s+1)|<2$ we are done. If $s+1 \in A$, we must have $s+1 \ge |b_0|$. For seeing that, suppose $s+1 < |b_0|$. So $s+1 \in X$. Therefore, since $s$ is the supremum, $s+1 \le s$ which is a contradiction. So we have $s+1 \ge |b_0|$. Since $s$ is the supremum and $|b_0|$ is an upper bound, we have $s \le |b_0|$. So, we have $s \le |b_0| \le s+1$. So $|s-|b_0|| \le 1$ and we are done.

Case 2:

Since $|b_0| \in A$, $b_0 \le 0$. This is because if $b_0 > 0$, then $|b_0|=b_0$. So we have $b_0 \in A$ which is a contradiction because by assumption, $A \cap B= \varnothing$. Put $$ Y=\{x \in A: x > b_0\}. $$ Either $1 \in A$ or $1 \in B$. If $1 \in B$, since $-1 \in A$ and $|-1-1| < 3$ we are done. So, we can assume, $1 \in A$. Therefore $1 \in Y$. So, $Y$ is inhabited. It is also bounded below. So, it has an infimum. The rest of the argument is similar to the case 1.