Any ordered field $F$ has characteristic $0$, so it contains a copy of $\mathbb{Z}$; by the universal property of the quotient field, the ring monomorphism $\mathbb{Z}\to F$ lifts to a monomorphism $\mathbb{Q}\to F$. We can identify $\mathbb{Q}$ with its image, so it's not restrictive to assume that $\mathbb{Q}\subseteq F$.
It's not really difficult: if $m/n\in\mathbb{Q}$, then we send it to
$$
\frac{f(m)}{f(n)}\in F
$$
where $f\colon \mathbb{Z}\to F$ is the (unique) monomorphism. Is this a field homomorphism? Just a check.
Now we come to the order. First of all, positive integers are positive in $(F,\prec)$: if $n>0$, then
$$
n=\underbrace{1+1+\dots+1}_{\text{$n$ times}}
$$
and therefore $0\prec n$. Conversely, if $n<0$, then
$$
n=-(\,\underbrace{\,1+1\dots+1}_{\text{$-n$ times}}\,)
$$
and so $n\prec0$.
Any element of $\mathbb{Q}$ can be represented as $m/n$ with $n>0$, because $a/b=(-a)/(-b)$, where $a,b\in F$, $b\ne0$. So, let $0\prec m/n$ in the ordering of $F$, with $n>0$. Then, by the properties of ordered fields,
$$
0\prec n\cdot\frac{m}{n}=m
$$
and therefore $m>0$. So a rational which is positive in $(F,\prec)$ is also positive in the usual order. A rational which is negative in $(F,\prec)$ is the opposite of a positive rational (in both orders).
Your confusion seems to arise because the Archimedes principle is stated in terms of $x,y$, and you have different $x,y$ in (b). Restate the Archimedean principle as:
(a) If $u,v$ are real numbers, with $u>0$ then there is a positive integer $k$ such that: $ku>v$.
(All I've done is change the variables, I hope.)
Now, $1$ is a real number, $y-x$ is a real number, and you've proven that $y-x>0$. So we know from (a) that if $u:=y-x$ and $v:=1$ that there is a positive integer, which we will call $n$, such that $(y-x)n>1$.
Similarly, since we know that $nx$ is a real number, and we know that $1$ is a real number and $1>0$, that from (a), setting $u:=1$ and $v:=nx$, that there is a positive integer we'll call $m_1$ such that $m_1\cdot 1 > nx$.
Finally, set $u:=1>0$ and $v:=-nx$ to show that there must be an $m_2$ so that $m_2\cdot 1>-nx$.
The last step is subtler, and doesn't use (a). Since $m_2>-nx$, $-m_2<nx$. So we know that $-m_2<m_1$.
Now, you need a property of the integers: If a non-empty set of integers has a lower bound, then it has a least element.
Take the set $S=\{m\in\mathbb Z: m> nx\}$. We know that $m_1\in S$, so $S$ is non-empty, and we know that $-m_2$ is a lower bound for $S$. So there is a least element $m\in S$. Then $m-1\notin S$, and therefore $m> nx$ and $m-1\leq nx$. So $m-1\leq nx< m$.
Best Answer
You mention in comments that your version of the Archimedean property is
Let $z\in\mathbb{R}$. Assume first that $z\gt 0$. Now, using the Archimedean property with $y=z$ and $x=1$, it follows that there exists $n\in\mathbb{N}$ such that $n\gt z$. Thus, the set $$\{ n\in\mathbb{N}\mid n\gt z\}$$ is nonempty. By the well-ordering principle, there is a least natural number $n_0$ such that $n_0\gt z$. Then $n_0\gt z$; if $n_0=1$, then $0\lt z\lt 1$ and $m=1$ works. If $n_0\gt 1$, then $n_0-1\in\mathbb{N}$, and minimality of $n_0$ means that $n_0-1\leq z$, Thus, $n_0-1\leq z\lt n_0$ and $m=n_0$ works again.
If $z=0$, take $m=1$.
If $z\lt 0$, then let $w=-z$. Then there exists, by the previous case, a nonnegative integer $k$ such that $k-1\leq w \lt k$. Therefore, $-k\lt z\leq 1-k$. If $z\lt 1-k$, then $m=1-k$ does the trick. If $z=1-k$, then $1-k\leq z\lt 2-k$, so $m=2-k$ works.