Given a set such that all its elements can be arranged in a particular order, with one to one correspondence, with the set of natural numbers.
How do we prove that there is no order on this set in which there are infinite elements between any two chosen elements?
Edit: I am looking for a proof which precludes the second ordering merely because the first one exists.
This is not true. For instance, if your set is $\mathbb{N}$, let $f:\mathbb{N}\to\mathbb{Q}$ be a bijection and define an order $<_f$ on $\mathbb{N}$ by $a<_f b$ iff $f(a)<f(b)$. There are then infinitely many elements between any two distinct elements $a$ and $b$ in this order on $\mathbb{N}$, since there are infinitely many rational numbers between $f(a)$ and $f(b)$.