Why isn't this a well ordering of $\{A\subseteq\mathbb N\mid A\text{ is infinite}\}$?

For each natural number $n$, let $A_n$ be the set $\mathbb{N}$ with the number $n$ removed from it. Then consider the collection $A = \{A_n \colon n \in \mathbb{N}\}$.

Then the set $A$ has no smallest element in your ordering, showing that it is not a well-ordering after all.

Proof: Note that for all $n$ we have $A_{n+1} < A_n$ since the first $n - 1$ elements of both sets are equal, and the $n$'th element of $A_{n+1}$ is $n$, which is smaller than the $n$'th element of $A_n$, which is $n + 1$.

So then $A_1>A_2>A_3> \dots $

And we find this is an ordering but the important thing here is that an ordering is called a WELL ordering if it is not only an ordering, but also has the additional property that every set has a smallest element.