I looked at the proof of Archimedean Property in several places and, in all of them, it is proven using the following structure (proof by contradiction), without much variation:
If $\space x \in \mathbb{R} \space,\space y \in \mathbb{R},$ and $x > 0$, then there is at least one natural number $n$ such that $nx > y$. $\bf{(*)}$
Proof:
Let $A$ be the set of all $nx$, where $n$ runs through the positive integers. $$A=\left\{nx \space| \space n \in \mathbb{N}\right\}$$ If $\bf{(*)}$ were false, which is equivalent to suppose that: $$nx \leq y, \space \forall n \in \mathbb{N}$$ then $y$ would be an upper bound of $A$. But then $A$ has a least upper bound in $\mathbb{R}$. Put $\alpha = \sup A$. Since $x > 0$, $\alpha - x < \alpha$, and $\alpha - x$ is not an upper bound of $A$. Hence $\alpha - x < mx$ for some positive integer $m$. But then $\alpha < (m+1)x \in A$, which is impossible, since $\alpha$ is an upper bound of $A$. Then, by contradiction, there exist an natural number n such that $nx > y$. $\tag*{$\blacksquare$}$
The following is a bit of what I found. All of these sources prove the property by contradiction:
Mathematical Analysis - Browder
A First Course in Analysis - Donald Yau
Principles of Mathematical Analysis - Rudin
A Course in Calculus and Real Analysis - Ghorpade & Limaye
Elementary Real Analysis - Thomson, J. Bruckner & A. Bruckner
So, my question is: How to prove this theorem directly, without using proof by contradiction?