I want to find interior of $\mathbb{Z}$ in $\mathbb{R}$. Here is the definition of $\mathring{A}$:
$$\mathring{A}=\{x\in \mathbb{R}\mid \exists r>0:(x-r,x+r)\subseteq A\}$$
My proof:
Suppose on contrary that $\mathring{\mathbb Z}\neq \emptyset$, so $\mathring{\mathbb Z}$ contains at least one element, say $y$. So, there is $r>0$ such that $(y-r,y+r)\subseteq \mathbb{Z}$. Since $r>0$, we can pick $N\in \mathbb{N}$ such that $r>1/N$; this is possible by Archimedean property. Then we have $y+\frac{1}{N}\in (y-r,y+r)$, but $\mathbb{Z}$ does not contain rational numbers, a contradiction.
Concerns:
What I am unsure about is if I used Archimedean property correctly, because it says "for all" $\varepsilon>0$, there is $N\in \mathbb{N}$ such that $\varepsilon>1/N$. Here, in my proof, the value $r>0$ is not arbitrarily given, only that it exists.