Show that $\mathbb N $ with the usual order is well ordered.
My attempt
Let $A\neq\emptyset$ a subset of $\mathbb N$. Since $A\subset \mathbb R$ is under-bounded by $0$, there is an infimum. Let denote $a=\inf A$. By property of the infimum, , there is a $x\in A$ such that $$a\leq x\leq a+1.$$ Since $a, x\in \mathbb N$, we necessarily have $x=a$, and thus $a\in\mathbb N$.
Is it correct ? It look to have a problem, the fact that I consider $\mathbb N$ as a subset of $\mathbb R$.
I think this is easiest: pick an $a\in A$. The set $$A'=\{a'\in A\mid a'\leq a\}$$ is finite and non-empty, and therefore has a smallest element $a_0$. Then $a_0$ will also be the smallest element of $A$.