Let $\mathbb{F}$ be an archimedean ordered field. Then $\mathbb{F}$ is isomorphic to a subfield of $\mathbb{R}$.
Let $x \in \mathbb{F}$, then define $S_x:=\{\frac{m}{n} \in \mathbb{Q}: \frac{m_{\mathbb{F}}}{n_{\mathbb{F}}}<x\}$.$\mathbb{F}$ being archimedean makes $S_x$ bounded in $\mathbb{R}$.
Define $\phi:\mathbb{F} \to \mathbb{R}$ such that $\phi(x)=\operatorname{lub}S_x$
Claim 1: $\phi$ is injective.
Let $x_1, x_2 \in \mathbb{F}$ with $x_1<x_2$ one can obtain an element of the form $\frac{m_{\mathbb{F}}}{n_\mathbb{F}}$ such that $x_1<\frac{m_{\mathbb{F}}}{n_\mathbb{F}}<x_2$.
Now how can I show that $\phi(x_1) \le \frac{m}{n}$?
Claim 2: $\phi(x+y)=\phi(x)+\phi(y)$ i.e $\operatorname{lub}S_{x+y}=\operatorname{lub}S_x+\operatorname{lub}S_y$
From the construction of $S_{a}$ it follows $S_x+S_y \subseteq S_{x+y}$ $\implies$ $\phi(x)+\phi(y) \le \phi(x+y)$
How do I show the reverse inequality?
Claim 3: $\phi(xy)=\phi(x)\phi(y)$ i.e $\operatorname{lub}S_{xy}=\operatorname{lub} S_{x}\operatorname{lub}S_y$
How do I do this?
I am writing @Paul K comment as an answer:
Claim 2: $\Phi(x+y)=\Phi(x)+\Phi(y)$ $\implies$ $\operatorname{lub}S_{x+y}=\operatorname{lub}S_x+\operatorname{lub}S_y$
By construction:
$S_x+S_y \subseteq S_{x+y} \Rightarrow \operatorname{lub}(S_x+S_y) \le \operatorname{lub}S_{x+y}$
$\Rightarrow \operatorname{lub}S_x+\operatorname{lub}S_y \le \operatorname{lub}S_{x+y}$ $\Rightarrow \phi(x)+\phi(y) \le \phi(x+y)$.
Let $q \in S_{x+y}$
$\Rightarrow$ $q_{\mathbb{F}}<x+y$.
Define $h=\left(x+y-q_{\mathbb{F}}\right)$.
Choose $p \in S_x$ and $t \in S_y$ with $x-h/2<p_\mathbb{F}<x$ and $y-h/2<t_{\mathbb{F}}<y$.
$\Rightarrow$ $x+y-h<p_{\mathbb{F}}+t_{\mathbb{F}}$ $\Rightarrow$ $q_{\mathbb{F}}<p_{\mathbb{F}}+t_{\mathbb{F}}=(p+t)_{\mathbb{F}}$.
So $q<p+t\le \Phi(x)+\Phi(y)$.
$\Phi(x)+\Phi(y)$ is an upperbound of $S_{x+y}$ $\Rightarrow \Phi(x+y)\le \Phi(x)+\Phi(y)$
Hence $\Phi(x+y)=\Phi(x)+\Phi(y)$.