Assuming that you agree with the existence of the largest element in $A$, the part of the proof in question follows as an easy consequence.
Let $n_0$ be the largest element in the set $A$. Then it satisfies the following properties:
- $n_0 \leq x$, since $n_0$ is an element of $A$.
- If $n < n_0$, then $n+1 \leq x$. This is clear from the inequality $n+1 \leq n_0 \leq x$.
- If $n > n_0$, then $n > x$. Indeed, assume $n > n_0$. Then $n$ is not a member of $A$, otherwise the maximality of $n_0$ is contradicted. Since $n \notin A$, we have $n > x$.
Since $n_0 + 1 > n_0$, we have $n_0 \leq x < n_0 + 1$, which proves the existence. Also, both 2 and 3 excludes the possibility that $n \leq x < n+1$ is satisfied for some $n$ other than $n_0$. This proves the uniqueness.
If you are questioning the existence of the largest element of $A$, you may exploit the well-ordering principle of $\Bbb{N}$ with a slight modification.