Let $F$ be an ordered field with the property that for all Dedekind cuts $(A,B)$ and all $n\in \mathbb{N}$ there are $a\in A$ and $b\in B$ such that $b-a< \tfrac{1}{n}$. I would like to conclude that $F$ is Archimedean but after numerous attempts I am not sure if this is true.
If $F$ is any ultrapower of $\mathbb{R}$ then in the cut determined by 0 you will find such elementes so we'd have to employ the assumption that it holds for all cuts, not only some particular ones.
Suppose not, then there are $0<\alpha<\beta$ in $F$ such that for all $n\in\mathbb{N}$, $n\cdot \alpha<\beta$.
Either $\alpha$ is infinitesimal and $\dfrac{1}{\alpha}>n\cdot 1$ for all $n\in \mathbb{N}$, or $\alpha$ is not infinitesimal, $m\cdot \alpha\geq 1$ and thus $\beta>n\cdot 1$ for all $n\in\mathbb{N}$.
Consider the set $A=\displaystyle{\bigcup_{n\in\mathbb{N}}}(-\infty,n\cdot 1)$. Then $A$ is a closed downward and $B=F\setminus A$ is non empty since either $\frac{1}{\alpha}\in B$ or $\beta\in B$. So, $(A,B)$ is a Dedekind cut.
By hypothesis, there are $a\in A,b\in B$ such that $b-a<1$, but this implies $$b<a+1\leq n\cdot 1 + 1=(n+1)\cdot 1$$ for some $n\in\mathbb{N}$, and we conclude that $b\in A$. obtaining a contradiction.