Working on proving the well ordering principle, I ran into this way to do it and I was hoping someone might read through it and tell me if I have any holes. I've looked over it a few times but I want to be sure.
Theorem — Each non-empty subset $S$ of $\Bbb N$ contains a smallest element.
Proof — Suppose that $S$ has no smallest element and define $T = \Bbb N \setminus S$. Suppose $\{1, 2, 3, ..., n\} \subset T$ and $n \in \Bbb N$. Further, suppose $\{1, 2, 3, ... , n\} \subset T \Rightarrow n + 1 \in T$. Then $n + 1 \notin S$ because it would be a least element of S. Since $1 \in T$ and $n + 1 \in T, T = \Bbb N$ by the induction axiom.
Since $T = \Bbb N \setminus S = \Bbb N, S = \varnothing$, which contradicts that $S$ is non-empty. Thus, each non-empty subset $S$ of $\Bbb N$ contains a smallest element. $\Box$
The only thing I can think of that might ruin my argument is supposing that $\{1, 2, 3, ... , n\} \subset T \Rightarrow n + 1 \in T$. I'm essentially saying that because induction works, the proof works.
But then again, this really shouldn't be a problem since the proof rests on the assumption that $S$ has no smallest element and therefore should maintain that property regardless of any subset of $\Bbb N$, including a set as fussily defined as $T$. Right?