So, to explain the title, I'm referring to the necessity of the axiom of choice in the existence of a well ordering on reals, or any uncountable set.
Now, while tweaking some sets, I came across this :
We start with the natural numbers, $N$. We take the power set of the naturals, $P(N)$.
Then we remove all the finite subsets of $N$ from $P(N)$. Let us call this new set $S$. This set is the set of all infinite subsets of $N$.
It is easy to show that $S$ has uncountable cardinality, same as that of real numbers. This is because the removal of finite subsets only removes a countable number of elements. (Haven't posted this deduction, for it is very easy, but I may post it if it is not so evident)
Now, we seek to find an ordering on the set $S$. Every set in this set is an infinite subset of natural numbers, so each of these sets are well ordered by the natural ordering of $N$.
Taking any two sets in $S$, say $A$, and $B$, we seek to order them by checking their elements lexicographically. We compare the first two elements in $A$, and $B$. Let them be $a_1$, and $b_1$ respectively. If $a_1 = b_1$, then we move on to the second elements in the sets, $a_2$, and $b_2$, and so on. If, at any point, $a_n < b_n$, then $A < B$, or if $b_n < a_n$, then $B < A$.
This order seems to be a well ordering of the uncountable infinity of reals. I don't seem to have invoked the axiom of choice anywhere in the construction of this set $S$.
So, why isn't this a well ordering on the uncountable of reals?