**Edit** Based on the comments above, I would like to rephrase my question. Suppose I change the recursive procedure as follows.
To proceed from $F_\alpha$ to $F_{\alpha+1}$ I perform the constructions below in the order shown.

- I first fill in the good/narrow gaps in $F_\alpha$ with cut points.
- For each bad/wide gap $(A,B)$, if there is a polynomial $p(x)$ which changes sign across the gap, I algebraically
extend $F_\alpha$ by a root $X_{(A,B)}$ of $p(x)$ contained within the gap (with degree of $p(x)$ minimal).
- For each bad/wide gap $(A,B)$, such that all non-zero polynomials in $F_\alpha$ have the same sign across the gap,
I add a transcendental element $X_{(A,B)}$. I declare a rational function $f(X_{(A,B)})$ over $F_\alpha$ to be
positive if the rational function $f(x)$ is positive across the gap
- Finally I add a transcendental element $X_\infty$ with a rational function $f(X_\infty)$ being declared
positive if $f(x)>0$ for all sufficiently large $x\in F_\alpha$

Do I get the surreals if I recurse over all ordinals?

It seems to me that steps (1) and (2) have the effect of eventually real closing each $F_\alpha$ at some later stage.

**Edit.** I overlooked Emil's last comment. If I understand him correctly, then my construction produces an $Ord$-saturated field.
Given sets $X<Y$, let $F_\alpha$ be the minimal field containing $X\cup Y$. Let
$$A=\{a\in F_\alpha\ |\ a\le x\in X\}$$
$$B=\{b\in F_\alpha\ |\ b\ge y\in Y\}$$
Then if $(A,B)$ is not a Dedekind cut of $F_\alpha$, then there is an element $u\in F_\alpha$ with
$X\subset A<u<B\supset Y$. If $(A,B)$ has a cut point $c$ in $F_\alpha$ then one of $c\pm\frac{1}{X_\infty}$
produces an appropriate $u\in F_{\alpha+1}$. Otherwise my construction explicitly produces such a $u\in F_{\alpha+1}$.

**Edit 2** I realize that I need to be a little more careful in formulating my construction. I need another transfinite recursion in the passage
from $F_\alpha$ to $F_{\alpha+1}$. First a couple of preliminary definitions.

Let $(A,B)$ be a Dedekind cut of an ordered field $F$. We say that $(A,B)$ is a gap if there is no cut point in $F$. We say that the
gap is narrow if for any $\epsilon>0$ in $F$, we can find $a\in A$, $b\in B$ with $b-a<\epsilon$. Otherwise we say the gap is wide.
We further classify the wide gaps as follows. If there is a polynomial $p(x)$ over $F$ which changes sign across the gap, we say the
gap is algebraic (relative to $F$). Otherwise we say the gap is transcendental (relative to $F$).

Now given an ordered field $F_\alpha$ I want to construct an ordered field $F_{\alpha+1}$ with the property that for any Dedekind
cut $(A,B)$ in $F_\alpha$ (gap or not)there is an element $u\in F_{\alpha+1}$, with $A<u<B$. I start by filling in all the narrow
gaps in $F_\alpha$ by taking the standard Dedekind completion of $F_\alpha$ as detailed in
this discussion.
I call the resulting field $F_\alpha^*$.

Now I start another transfinite recursion. I begin by well-ordering all the wide gaps $(A_\beta,B_\beta)$ in $F_{\alpha}$. I define
$F_{\alpha,0}=F_\alpha^*$. Having defined $F_{\alpha,\beta}$, I define $F_{\alpha,\beta+1}$ as follows.

Case 1. If there is already an element $u\in F_{\alpha,\beta}$ with $A_\beta<u<B_\beta$, then I take $F_{\alpha,\beta+1}= F_{\alpha,\beta}$.

If we are not in Case 1, then we can extend the wide gap $(A_\beta,B_\beta)$ in $F_\alpha$ to a wide gap $(A'_\beta,B'_\beta)$
in $F_{\alpha,\beta}$.

Case 2. If $(A'_\beta,B'_\beta)$ is algebraic relative to $F_{\alpha,\beta}$, then I take $F_{\alpha,\beta+1}$ to be an algebraic
extension of $F_{\alpha,\beta}$ obtained by adjoining a root $X_{(A'_\beta,B'_\beta)}$ of an irreducible polynomial $p(x)$ which
changes sign across the gap. We order $F_{\alpha,\beta+1}$ so that $X_{(A'_\beta,B'_\beta)}$ lies in the gap.

Case 3. If $(A'_\beta,B'_\beta)$ is transcendental relative to $F_{\alpha,\beta}$, then I take $F_{\alpha,\beta+1}$ to be a transcendental
extension of $F_{\alpha,\beta}$ obtained by adding a transcendental element $X_{(A'_\beta,B'_\beta)}$ with a rational function
$f(X_{(A'_\beta,B'_\beta)})$ declared positive if the corresponding function $f:F_{\alpha,\beta}\to F_{\alpha,\beta}$ is positive across the
gap.

After completing the induction $F_{\alpha,\beta}$ over $\beta$, I obtain a field $\widehat{F}_\alpha$ which contains elements spanning
all gaps in $F_\alpha$. Finally I add a transcendental element $X_\infty$ to $\widehat{F}_\alpha$ with a rational function $f_(X_\infty)$
declared positive if the corresponding function $f:\widehat{F}_\alpha\to \widehat{F}_\alpha$ is positive for sufficiently large values
of the argument. I call the resulting field $F_{\alpha+1}$. This last step has the effect of adding elements between Dedekind cuts
in $F_\alpha$ which are not gaps, but rather have cut points.