I'm claiming that this proof is using the fact that $u < m+1$, which isn't valid. Because for any finite set of positive integers $K$, $\sup K = \max K$. Which means that it would be impossible for the supremum of a set of positive integers to be less than another integer in the same set. That is, the number $m$ in the author's proof must be the next integer less than the supremum.
Proof
Firstly, for all $k \in K$, $k \leq \max K$. So $\max K$ is an upperbound for $K$.
Now, let $v \in \mathbb{R}$ be any number less than $\max K$. Define $\epsilon = \max K - v >0$. Choose $s_\epsilon = \frac{\max K+v}{2}$. Then
$$\max K - \epsilon = v < \frac{\max K+v}{2} = s_\epsilon $$
Proving that $\sup K = \max K$. That means the author's supremum $u$ in the proof is the maximum integer of the nonempty set $\mathbb{N}$. Subtracting $1$ from $u$ would provide the next lowest integer. There would indeed exist $m \in \mathbb{N}$ s.t. $u -1 < m$ but that $m$ would have to equal $\max{\mathbb{N}}$. So, in conclusion, $u \nless m +1$.
Please let me know what you think.