$\DeclareMathOperator{\cof}{cof}$ Let $F$ be an ordered field in which any strictly increasing bounded map $\cof(F) \rightarrow F$ is Cauchy. Let's say that $\cof(F) > \omega_0$, otherwise $F$ is archimedean.
Does the real closure $\widetilde{F}$ of $F$ have the same property?
Note that if a bounded strictly increasing $\cof(F)$-sequence $u$ in $\widetilde{F}$ has a uniformly bounded family of monic anihilator polynomials over $F$, then there is a subsequence $v$ of $u$ for which all those polynomials have the same degree and number of roots in $\widetilde{F}$, and one can prove by induction on the degree that all those coefficients have Cauchy subsequences. A suitable application of the theorem of continuity of roots for real closed fields yields that $v$ has a Cauchy subsequence and is therefore Cauchy.
However, since $u$ is a $\cof(F)$-sequence, there does not seem to be a way to arrange that it has a family of uniformly bounded anihilator polynomials (see the anwser to this previous MathOverflow question of mine). So this might not be a good way to prove this if it's true.
Any ideas of different approach or of ways to reduce to when such a family exists would be appreciated.