On page 19 of Principles of Mathematical Analysis by Walter Rudin ("Baby Rudin"), Rudin proves the field axioms of addition for Dedekind cuts ("Step 4" of the Ch.1 Appendix; the entire text can be found here).
In the process of proving that 0*$\subseteq{\alpha+\beta}$, Rudin states that for a positive $w\in{\mathbb{Q}}$, "there is an integer $n$ such that $nw\in{\alpha}$ but $(n+1)w\notin{\alpha}$," and adds "Note that this depends on the fact that $\mathbb{Q}$ has the archimedean property!"
I see that the Archimedean Property implies that $\exists{n_1\in\mathbb{Z}}$ such that $n_1w\in\alpha$, since we can choose $p\in\alpha$ and $n_1\in\mathbb{Z}$ such that $n_1w<{p}$. Similarly, given a $q\notin\alpha$, we can pick $n_2\in\mathbb{Z}$ such that $n_2w>q$, which implies $n_2w\notin\alpha$. But what makes Rudin so sure we can pick $n_1$ and $n_2$ such that $n_1+1=n_2$?
The way I would think to do this (were this not in the middle of a proof about how the Dedekind cuts are a field satisfying the Least Upper Bound Property) would be to use the Least Upper Bound Property of $\mathbb{R}$, of which $\mathbb{Q}$ is a subset. The set $S=\{nw\in\alpha\mid{n\in\mathbb{Z}}\}$ is unempty by the Arch. Prop. and bounded above by, for example, $q$ from the previous paragraph. Therefore, but the L.U.B. Property, $A=\sup{S}$ exists. Letting $\epsilon=\frac{w}{2}$, there is at most one element $s\in{S}$ such that $A-\epsilon<s\leq{A}$. Such an $s$ must exist by the Approximation Property of Sup. If we had $s\neq{A}$, we could let $\epsilon'=|A-s|$. Then the only possible value $s'$ such that $A-\epsilon'<s'\leq{A}$ is A itself, so by the Approximation Property of Sup, we must have that $A\in{S}$. In that case, $A=nw\in\alpha$ for some $n\in\mathbb{Z}$, but $(n+1)w\notin\alpha$. So the claim can be proved this way.
But can we prove it without using the L.U.B. Property of $\mathbb{R}$? Seems a little strange to do so when the point of this whole section of the book is to construct the field $\mathbb{R}$.