Prove: if for all $x,y \in A \subset \Bbb R$, $|x-y|<1$, then $\sup A-\inf A \le 1$

Question

Consider the fact: if for all $x,y \in A \subset \Bbb R$, $|x-y|<1$, then $\sup A-\inf A \le 1$.

It's not difficult to show that if one supposes that $ \sup A-\inf A > 1$, then there exist $x,y \in A$ such that $|x-y| \ge 1$, and that, therefore, the proposition above cannot be false.

Is it in any way imaginable to give a direct proof of this proposition?

Answer 1

\begin{align} \forall x,y\in A, |x-y|<1&\implies \sup_{x,y\in A}|x-y|\leq 1\\ &\implies \text{diam(A)}\leq 1\\ &\implies \sup(A)-\inf(A)\leq 1, \end{align} because if $A\subset \mathbb R$, then $$\sup(A)-\inf(A)\leq \text{diam}(A).$$

Answer 2

$-1<x-y<1 \ \forall \ x,y\in A$ so $x<1+y,\ \forall \ x,y\in A \Rightarrow \sup A\leq 1+y ,\ \forall \ y\in A \Rightarrow y\ge \sup A -1, \forall \ y\in A\Rightarrow \inf A \ge \sup A-1$