An answer using Peano's axioms is provided here: Use an induction argument to prove that for any natural number $n$, the interval $(n,n+1)$ does not contain any natural number.
I'm working through royden's text and it didn't provide those axioms. So I'm wondering if this can be proven without those axiom. The preliminaries given are the algebra of sets, equivalence relations, zorn's lemma, axiom of choice, the field properties, definition of an inductive set, every subset of $\Bbb{N}$ has a least element, archimedean property, and $\Bbb{R}$ is dense. Particularly, I used this definition of an inductive set
A set $E$ of real numbers is inductive provided that it contains $1$ and if $x\in E$, then $x+1\in E$
Proof: (Base case: n=0) Considering $(0,1)$ there cannot be any natural number in the interval since $1$ is the least natural number. (I'm not quite sure here. Royden did mention in a proof that $\Bbb{N}$ is bounded below by $1$ so I thought I could argue this way. Could I?)
(Inductive Case:) Suppose $\forall n\in\Bbb{N}$, $(n, n+1)\cap\Bbb{N}=\emptyset$. Since $\Bbb{N}$ is inductive, $n+1,n+2\in\Bbb{N}$. Also, we note that $\forall q\in(n,n+1)$, then $n<q<n+1 \Leftrightarrow n+1<q+1<n+2$. Keep in mind that $(n,n+1)\cap\Bbb{N}=\emptyset\Rightarrow q\notin\Bbb{N}$. By definition of an inductive set, $q+1\in (n+1,n+2)\Leftrightarrow q\in\Bbb{N}$. Since $q\notin\Bbb{N}$, then $q+1\notin\Bbb{N}$ and $(n+1,n+2)\cap\Bbb{N}=\emptyset$. (I interpreted the definition as an if and only if statement since that's what I was told when dealing with definitions.)
The following uses these properties of $\Bbb N$:
Let $$ S=\{\,m\in\Bbb N\mid \exists n\in\Bbb N\colon n<m<n+1\,\}.$$ The claim we want to prove is equivalent to $S=\emptyset$. So assume $S$ is not empty and let $m=\min S$. Then there exists $n\in\Bbb N$ with $n<m<n+1$. From $m>n\ge 1$, we see that $m-1\in \Bbb N$. Then from $n-1>m-1\ge1$, we see that $n-1\in\Bbb N$. Then $n'<m'<n'+1$ with $n'=n-1$, $m'=m-1$ shows $m'\in S$, contradicting minimality of $m$. $\square$