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 $m-1\leq nx< m$.
I'm not sure whether Rudin was implicitly appealing to the well-ordering principle, but it is possible to interpret what he's written in a way that doesn't use it.
Namely, the set $S$ of integers that are $\leq nx$ is nonempty, by the existence of $-m_2$, which belongs to the set, and bounded above, by the existence of $m_1$, which is an upper bound for it, and does not belong to it. Therefore $S$ must have a least upper bound $p$. $p$ must belong to $S$, for otherwise $S$ would have to have two elements strictly between $p-1$ and $p$, which is impossible since the elements of $S$ are integers. Then let $m = p+1$.
The inequality $-m_2 \leq m$ follows from the fact that $p = m-1$ is the largest element of $S$, and $-m_2 \in S$. The inequality $m \leq m_1$ results from $m-1 = p \leq nx < m_1$. And the inequalities $m-1 \leq nx < m$ or, equivalently, $p \leq nx < p+1$, are a consequence of the fact that $p$ is the largest element of $S$.
I would say that this proof uses a number of unstated assumptions about the integers, their basic properties, and their relationship to the real numbers (for example, the fact that the order relation is the same, or the fact that a successor integer is obtained by adding $1$ in $\mathbf{R}$). This is unavoidable, because the integers haven't been defined. However, it does not appear that an appeal to the well-ordering principle is necessary. The proof is logically deficient only to the extent that the integers are assumed to have these unstated properties.
I don't believe the proof can be substantially improved without formalizing what the integers are, perhaps defined as a subset of $\mathbf{R}$ with certain properties, if one wishes to retain the axioms for $\mathbf{R}$ as the basis for the exposition.
Best Answer
The integers $m_1$ and $m_2$ serve to bound $nx$ between two integers. The set $$=\{-m_2+1,-m_2+2,\ldots,m_1\}$$ is a finite set of integers, so we can choose the smallest member $m$ of this set such that $nx<m$.
If we knew that $nx$ was positive, we wouldn’t need $m_2$: we could just choose the smallest positive integer $m$ such that $nx<m$, since every non-empty set of positive integers has a least element. In fact, we can use that well-ordering principle directly, once we have $m_1$ and $m_2$. Let
$$M=\{m\in\Bbb Z^+:m-m_2>nx\}\;.$$
Then $M\ne\varnothing$, since $m_1+m_2\in M$, so $M$ has a least element, say $k$. Let $m=k-m_2$. Then $m>nx$. However, $k-1\notin M$, so $m-1=k-1-m_2\not>nx$, i.e., $m-1\le nx$. But note that I needed both $m_1$ and $m_2$ to carry out this argument: $m_1$ is needed to ensure that there is at least one integer that’s big enough to exceed $nx$, and $m_2$ is needed to ensure that not every integer is big enough.