Let $S$ be a nonempty set of natural numbers. We show that there is an $a\in S$ such that for all $b\in S$ $a\leq b$ Suppose not, then $$(*)\ \ \ \ \ \forall a\in S,\exists b\in S\ \ b<a.$$ We will use the principle of finite induction (the strong form) to show that $S$ is empty, a contradition.

Fix any natural number $n$ and suppose that for all natural numbers $m<n$ $m\in \mathbb{N}\setminus S$ If $n\in S$ then (*) implies that there is an element $b\in S$ such that $b< n$ This would be incompatible with the assumption that for all natural numbers $m<n$ $m\in \mathbb{N}\setminus S$ Hence, we conclude that $n$ is not in $S$

Therefore, by induction, no natural number is a member of $S$ The set is empty.