How can I show the incompleteness of the irrational numbers?

Consider all irrational numbers smaller than $0$. This set doesn't have a supremum (in the irrational numbers).

If you want an explicit sequence in $\mathbb{I}$ which limits to a rational, try $a_n = -\sqrt{\frac{1}{n}}$ for $n$ non-square.

What Stirling's formula asserts is a counterexample:

$$\lim_{n\to\infty} \frac{n!}{\sqrt{2\pi n}\,\Bigl(\dfrac{n}{\mathrm e}\Bigr)^{\!n}}=1. $$