One cannot prove the Archimedean property for the reals by appealing only to first-order algebraic truths of the ordered real field and the subring of the integers sitting inside it. The reason is that those statements are all first-order expressible in the structure $\langle\mathbb{R},{+},{\cdot},0,1,\lt,\mathbb{Z}\rangle$, and there are nonstandard models $\langle\mathbb{R}^\ast,{+},{\cdot},0,1,\lt,\mathbb{Z}^\ast\rangle$, which are not Archimedean and yet which satisfy exactly the same first-order truths as the standard model. In this sense, there cannot be a truly elementary proof of the Archimedean property.
What this observation shows is that the Archimedean property is fundamentally a second-order property of the reals, and to establish it one must appeal to the fact that one is using the actual standard integers instead of merely some first-order property of the integers and how they relate to the reals.
Your argument, which is fine, does this by appealing to the explicit constructions of the reals, say, as Dedekind cuts, rather than to its characterization by the completeness property.
So while I agree that there are other proofs of the Archimedean property that don't appeal to completeness---and as you point out there certainly are incomplete Archimedean ordered fields---nevertheless I don't mind the proof from completeness, since this is a defining property characterizing the reals, and the proof from completeness is both easy and clear.
