Natural number between two real numbers $y,x$ such that $y-x > 1$

Prove that for $x,y \in \mathbb{R}$ such that $y-x>1$ there is a natural number $n$ such that $x < n < y$.

Consider the following set:

$$S := \{n \in \mathbb{N} | x < n < y\}$$

Suppose $S = \emptyset$, therefore $\forall n \in \mathbb{N}$$n \leq x$ or $n \geq y$. Let $n_1$ be the first natural number before $x$. Therefore $n_1 < x < n_1 + 1$.

Now we'll show that $n_1+1 < y$. Suppose not. Hence $n_1<x<y<n_1+1$. Which implies that

$$ 1 = n_1 + 1 - n_1 > y-x > 1$$

Which is a contradiction, therefore, $x<n_1+1<y \rightarrow (n_1+1) \in S$.

Here's what I'm thinking about my proof: I didn't use the fact that $x,y$ are real numbers... Maybe I wasn't supposed to prove it that way? Can someone please point out another proof if mine is not correct for that case? I don't know, it just feels weird to prove something and don't use all our premises.

Thanks!