First the book defines the reals as a set with two operations that satisfy the field and order axioms. Then, a set $S \subseteq \mathbb{R}$ is said to be inductive if $1 \in S$ and if $x \in S \implies (x+1) \in S$. Then the natural numbers are defined as the intersection of all inductive subsets of $\mathbb{R}$. To show that there is no $n \in \mathbb{N}$ such that there is $k \in \mathbb{N}$ such that $n < k < n+1$, the books suggests showing that the set $A_n \cup B_n$ is inductive where $A_n = \{ x \in \mathbb{N} | x \leq n \}$ and $B_n = \{ x \in \mathbb{N} | x \geq n+1 \}$. And this would indeed prove the desired result, I did it by induction.
I was wondering, couldn't this be a lot easier? and I thought, first we observe that the natural numbers must be exactly $1$ and its successors, this is because the set of $1$ and its successors form an inductive set thus $\mathbb{N} \subseteq\{1, 2, 3, ... \}$ and also $1$ and its successors belong on every inductive set therefore $\{1, 2, 3, ... \} \subseteq \mathbb{N}$. From here notice that if $n, m \in \mathbb{N}$ then $n < m$ iff $m$ is a higher succesor of $1$ than $n$. Then suppose for contradiction that $n, k \in \mathbb{N}$ such that $n < k < n+1$. Then $k$ is a higher successor of $1$ than $n$, that is, it is $n$ plus $1$ added a certain amount of times, but the least amount of times we can add $1$ is one and $k$ is a lesser succesor of $1$ than $n+1$, a contradiction.
Is this proof good? If not, then what is it lacking? I feel like there's no need to do anything overly formal to prove this result, once you prove that the intersection of all inductive sets must be $1$ and its successors, then the theorem should be pretty obvious I feel like, but maybe I am missing something and we do need more elaborate proofs for some reason, what do you think?