"Call "k" the biggest natural number, which is smaller than s."
You have no reason to believe such a thing exists.
But you can prove it.
If $S \subset \mathbb N$ that is bounded above then $s \sup S$ exists.
Let $j = s-1$. Then $j$ is not a upper bound of $S$. So there exists a natural $k$ so that $j=s-1 < k \le s$.
Thus $s = j+1 < k+1 \le s+1$ and $k+1$ is not in $S$. Nor are any natural numbers that are greater than $k$. So $k$ is the greatest natural number in $S$.
So every upper bounded set of natural numbers will have a greatest element.
But this also proves that for any upper bounded set $S$ of natural numbers there exists natural numbers not in $S$ (namely all the natural numbers larger than $k$).
Thus we no longer have anything left to prove. (No upper bound set of natural numbers contain all natural numbers, so $\mathbb N$ itself which does contain them all, can not be bounded above.)
So actually, "every upper bounded set of natural numbers has a maximum element" and "the set of natural numbers are not bounded above" are equivalent statements both with the same proof.