Axiom of Choice and the cardinality of the reals

Your error is thinking that "well-ordered and infinite" means "bijectable with $\mathbb{N}$". Your argument is not even enough to give a bijection between $\mathbb{N}$ and the following well-ordering of the integers: order the nonnegative integers in the usual way; make every negative number larger than any nonnegative number, and compare negative numbers by comparing their absolute value. That is, the well ordering $$0, 1, 2, 3,\ldots, n,\ldots ; -1, -2, -3, \ldots, -n, \ldots$$ where ";" means that $-1$ is larger than any nonnegative integer. This type of order is called $\omega+\omega$, because it is essentially two copies of $\mathbb{N}$, one placed after the other ($\omega$ is the ordinal name of the well-order of the natural numbers). This is still countable, of course, but you can probably see already that your argument about well-ordering the reals to get a bijection with $\mathbb{N}$ is already in serious trouble: you have no warrant for assuming that it will actually "hit" every real number (and in fact, it won't).

Added: Just for completeness: to show this is a well ordering of $\mathbb{Z}$, let $A$ be any nonempty subset of $\mathbb{Z}$. If $A\cap\mathbb{N}$ is nonempty, then the least element of $A$ is the least element $\mathbf{a}$ of $A\cap\mathbb{N}$ (my naturals include $0$, by the by), since given any $a\in A$, if $a\in\mathbb{N}$ then by definition of $\mathbf{a}$ we have $\mathbf{a}\leq a$. And if $a$ is negative, then since $\mathbf{a}$ is nonnegative we have $\mathbf{a}\leq a$. Thus, $\mathbf{a}$ is the least element of $A$. If, on the other hand, we have $A\cap\mathbb{N}=\emptyset$, then that means that $A$ consists only of negative numbers. Let $B=\{ |a|\mid a\in A\}$. Then $B\subseteq\mathbb{N}$ and is nonempty, so it has a least element $\mathbf{b}$. Then $\mathbf{a}=-\mathbf{b}\in A$ is the least element of $A$, since given any $a\in A$, we have that $a$ is negative by assumption and so that $|\mathbf{a}| = \mathbf{b}\leq |a|$; since this is how we compare negative numbers in this order, we have that $\mathbf{a}$ is less than or equal to $a$, hence $\mathbf{a}$ is the least element of $A$, as claimed.

Well orderings can be much longer than $|\mathbb{N}|$. There certainly is an nth real in the well-order, but there are reals with much higher ordinals as well. One form of AC is that any set can be well-ordered, but that does not imply that all infinite sets have the same cardinality.