[Math] Is this proof of the Archimedean Property valid

Question

Archimedean property: The set of natural numbers $\mathbb{N}$ is not bounded above.

Proof :

Suppose $\mathbb{N}$ is bounded above. Then, by the supremum property, there exits a lowest upper bound "$s$" for all $n \in\mathbb{N}$. Call "$k$" the biggest natural number, which is smaller than $s$. Then $k+1>s$ and since $k+1$ is a natural number, $s$ is not an upper bound of $\mathbb{N}$. Thus, we have reached a contradiction and can conclude that $\mathbb{N}$ is not bounded above.

Answer 1

Assume N subset R is bounded above.
So there is a least upper bound of N. Call it s.
If n in N, then n + 1 in N. Hence n + 1 <= s, n <= s - 1.
Thus s - 1 is a smaller upper bound of N, a contradiction.