Well-ordering proof

What is wrong with this "proof" of the following statement?

For every positive integer $n,$ the number $n^2+n+1$ is even.

Proof:

Let $S$ be the subset of positive integers $n$ for which $n^2+n+1$ is odd. Assume $S$ is nonempty.

Let $m$ be its smallest element.

Then $m-1 \notin S,$ so $(m-1)^2+(m-1)+1$ is even.

But $(m-1)^2+(m-1)+1 = m^2-m+1 = (m^2+m+1)-2m,$ so $m^2+m+1$ equals $\big((m-1)^2+(m-1)+1\big)+2m,$ which is a sum of two even numbers, which is even.

So $m \notin S;$ which is a contradiction. Therefore, $S$ is empty, and the result follows.