Let $F$ be an ordered field. Let $A$ be a convex subring of $F$. $A$ is a valuation ring on $F$, so a local ring. Let $m_A$ denote its maximal ideal.
As a general result of valuation-theory, we know that for any subfield $C$ of $F$ contained in $A$ that is maximal for inclusion with respect to these conditions, the residue field $F_A:= A / m_A$ is algebraic over the image $C_A$ of $C$ under the residue map $A \rightarrow A /m_A$. Moreover $C$ is algebraically closed relatively to $F$.
Now, in this more specific case, $F_A$ is ordered by $a + m_A < a' + m_A \Longleftrightarrow a < a'$.
If $F$ is real closed, then so is $C_A$, and $C_A = F_A$.
In general, $C_A$ may be a proper subfield of $F_A$; for instance, if $F$ is a ultrapower of $\mathbb{Q}$ modulo some free ultrafilter on $\mathbb{N}$ and $A$ is the ring of finite and infinitesimal elements in it, then $F_A \cong \mathbb{R}$ while any maximal subfield of $A$ may contain no algebraic irrationnal number.
Is $C_A$ always dense in $F_A$?
The answer turns out to be yes.
In fact, a little more generally, if $F$ is an ordered field, $F'$ is a proper extension of $F$ and $a \in F' \smallsetminus F$ is such that there is $\varepsilon \in F^{>0}$ such that $\forall x \in F(|x-a| > \varepsilon)$, then there is $b \in F' \smallsetminus F$ such that $\forall x \in F(x < a \longleftrightarrow x < b)$ and that $F(b)$ is a cofinal extension of $F$.
Applying this to $C_A$ assuming it is not dense yields such an element $a$ for $C$ (in $A$) which contradicts the maximality of $C$ as a subfield of $A$.