Density of linear order where $|A| = 1$

Question

Let $\le$ mean a linear order in a non-empty, finite set $A$. I need to prove that this order is dense only when $|A| = 1.$

I understand that when I assume that $|A| = 1$, then I have: $$\forall x,y\in A\Big((x \lt y) \rightarrow \exists z (x<z<y)\Big)$$ and that's true.

But I don't quite understand the case when $|A| \ge 2$.

Because then I have: $$\forall x,y\in A\big(\exists z\neq x,(y\mid x<z<y)\big). $$ And that's for some reason contradiction.

And I really want to understand and learn this.

Answer 1

List in order, the elements a,b,.. n.
If there is more than one element, pick a,b, the first two.
Since there is no x such that a < x < b, the order is not dense.