I've seen two different proofs of Archimedean property:
$\boxed{(1)}$
For every $r∈ℝ$ there exist $n∈ℕ^+$, such that: $n>r$
Proof:
There are two cases to consider, the case where $r\le0$ and the case where $r>0$.
For the first one let $n∈ℕ^+$ and the result follows, for the second one let $n\le r$, then clearly the set of all these $n$'s is not empty , also since the set $ℕ$ is bounded above by $r$ and $ℕ⊂ℝ$ using Completeness axiom implies sup$ℕ$ exist, denote it by $α$, then clearly $α-1$ is not an upper bound for the set $ℕ$, which means: $$∃ \thinspace m∈ℕ:α-1< m$$ or equivalently :$$∃ \thinspace m∈ℕ:α< (m+1) ∈ℕ$$ which is a contradiction based on the definition of sup$ℕ=α$, hence the property is true.
$\boxed{(2)}$
For every real number $y$ there exist $ x∈ℝ^+$ and $n∈ℕ^+$ such that: $nx>y$
Proof:
There are two cases to consider, the case where $y\le0$ and the case where $y>0$.
For the first one let $n∈ℕ^+$ and the result follows, for the second one let $nx\le y$, then clearly the set of all these $ny$'s denoted $A$ is not empty , also since the set $A$ is bounded above by $y$ and $A⊂ℝ$, using Completeness axiom implies sup$A$ exist, denote it by $α$ ,then since $ x∈ℝ^+$ the inequality $α-x\leα$ holds , it can be concluded that $α-x$ is not an upper bound for the set $A$, means: $$∃ \thinspace m x∈A:α-x< mx$$ or equivalently :$$∃ \thinspace m x∈A:α< (1+m)x ∈A$$ which is a contradiction based on the definition of sup$A=α$, hence the property is true.
I think the first proof is actually a special case of the second one when $x=1$, also in the first proof we claim that for every real number there exist a natural number greater than that, but in the second proof since $ x∈ℝ^+$ and $n∈ℕ^+$ , implies $nx∈ℝ^+$ , here we actually claim that for every real number there exist another real number greater than that, this real number is not necessarily natural (non-zero) , so based on these two reasons I think the second case is a generalized case of the other one, but I'm not sure if I'm right or not.