Well-Ordering Principle "proof" Theorem. Well-Ordering Principle.
Every non-empty subset of natural numbers has a least element.
I have seen some proofs for the theorem, but is very "complex"proof really needed here?
My attempt of proof:
Let $$ D \subset \mathbb{N}= \left\{ 1, \ 2, \ 3, \ \dots \right\} $$
be an arbitrary non-empty subset of natural numbers. Therefore it has at least one element $$  n \in D.$$
Consider the finite set $$ \left\{1, \ \dots, \ n \right\}.$$
We check which of those natural numbers are elements of D. Then we choose the smallest one of those. There we have the least element.

aschepler's comment points out the problem exactly. First you construct the set $$D\cap \{ 1,2,\ldots n\}$$ and then you consider the smallest element of this set...
Except that without the well-ordering principle, you cant be sure that it has a smallest element!
It might be instructive to consider what happens in your argument if you try to apply it to prove that $\Bbb Z$ is well-ordered.
