How to prove rigorously that, for any natural number $n$, the open interval $(n, n+1)$ contains no natural number?

Question

A subset $E$ of $\mathbb{R}$, the set of real numbers, is said to be inductive if $1 \in E$, and for every $x \in E$ the number $x+1 \in E$ also.

The set $\mathbb{N}$ of natural numbers is defined to be the intersection of all the inductive sets of real numbers.

Since the sets $\{ x \in \mathbb{R} | x > 0 \}$ and $\{ x \in \mathbb{R} | x \geq 1 \}$ are inductive, therefore we can conclude that all the natural numbers are $\geq 1$.

Now using this machinery how can we (use induction to) prove the following statement?

For any natural number $n$, there exists no natural number in the open interval $(n, n+1)$.

Answer 1

Since it is an open interval, any $x$ in $(n, n+1)$ satisfies $x > n$ and $x < n + 1$.

We can use these two inequalities to show that $x$ cannot be an integer. Assume for contradiction that $x$ is an integer:

$x > n$ is equivalent to $x \ge n + 1$ but this contradicts $x < n + 1$. QED.

Answer 2

I'll give it a naive try.

Let's attempt to prove it by induction, and lets set $n=0$ to prove our base case. Since all natural numbers are bigger than $1$ (or are $1$), we know that there are no natural numbers in $(0,1)$. Base case proved.

Now let's set $n=k+1$, instead of $n=k$. Then we need to show that $(k+1,(k+1)+1)$ reduces to $(k,k+1)$ somehow. Now, for every element in $x\in(k+1,(k+1)+1)$, really $x\in(k,k+1)+1$, making the union of all sets of the form $(n,n+1)$ inductive if $1\in{(0,1),(1,2),(2,3),...}$, which it's not. So because $1$ is not in this set, $1+k$ for $k=0,1,2,...$ is also not in this set by the definition of inductive, so we cannot get the natural numbers from these sets by the definition of a natural number.