Are archimedean subextensions of ordered fields dense?

Let $E$ be the real closure of $\mathbb{Q}(x, y) = (\mathbb{Q}(x))(y)$, with order given by $x > \mathbb{Q}$ and$y > \mathbb{Q}(x)$. In other words, positivity on $\mathbb{Q}(x, y)$ is determined first by degree in $y$, and then by degree in $x$. Then let $F$ be the real closure of $\mathbb{Q}(y)$.

First, we prove that $E$ is $F$-archimedean. Let $e \in E$. There is some $e' \in \mathbb{Q}(x, y)$ with $e' > e > -e'$. Then $e'$ has degree $n$ in $y$ for some $y$; let $f = y^{n + 1}$. By the order on $\mathbb{Q}(x, y)$, we have $f > e' > e > -e' > -f$. Therefore $E$ is $F$-archimedean.

On the other hand, clearly, there is no element of $F$ between $x$ and $x + 1$.

Let $F$ be any real-closed field of uncountable cofinality. That is, every countable subset of $F$ is bounded. One can make such a field in a process of $\omega_1$-many field extensions; alternatively, the ultrapower of $\mathbb{R}$ by nonprincipal ultrafilter on $\omega$ also has uncountable cofinality.

Let $E=F^\omega/\mu$ be an ultrapower of $F$ by a nonprincipal ultrafilter $\mu$ on $\omega$. We may identify $F$ with its canonical copy in $E$ using equivalence classes of constant functions $x\mapsto [c_x]_\mu$.

Since every function from $\omega$ to $F$ is bounded by a constant function, it follows that $E$ is $F$-archimedean. But $F$ is not dense in $E$, since there are no constant functions between $[\text{id}]_\mu$ and $[\text{id}+1]_\mu$, where $\text{id}:n\mapsto n$, viewing $\omega\subset F$.

This is similar to user44191's answer but I want to put the emphasis on the fact that there is no reason that the cofinality of $F$ in $E$ (which is what ou call [$E$ is $F$-archimedean]) imply the density of $F$ in $E$.

Indeed, if $F$ is any non-archimedean ordered field, then the field $F(t)$ can be equipped with an order where $t$ is positive infinite but smaller than any positive infinite element of $F$. Thus $F$ is cofinal in $F(t)$ but not dense in it since no element of $F$ is close to $t$. To do so, say that a fraction $\frac{P(t)}{Q(t)}$ is positive if $P(t)$ and $Q(t)$ have the same sign, where the sign of a polynomial $R(t)$ is positive if $R(s)$ is positive for sufficiently large finite element $s$ of $F$.

This is a special case of filling a cut in an ordered field using a simple extension.