Set theory says (using the Axiom of Choice) that every set can be well ordered; how can that be, when the...

The confusion arises because of the similarity of terminology.

The field of complex numbers is NOT an "ordered field" in the sense
that the field of real numbers is. By this, we mean that there is not
a relation < for which the following properties hold:

        (1) for any a,b precisely one of the following holds:
                (i)        a<b
                (ii)        b<a
                (iii)        a=b
        (2) if a<b and 0<c then ac < bc
        (3) a<b and c<d => a+c < b+d

Note that any relation satisfying these conditions must be transitive:

        a<b & b<c => a<c

Further, note that for every nonzero a, 0<a*a, since either x<0 or 0<x
for every nonzero x, and thus either 0<a or 0<(-a). Since both square
to give a*a, condition (2) implies that 0<a*a. Next, note that i*i is
a square, but i*i = -1. Since 0<1, we note that 0<(-1) also, and get a
contradiction by adding:

        0<1 & 0<(-1) => 0+0 < 1 + (-1)
                thus:         0  <   0  (but evidently, 0=0: contradiction)

Thus the field of complex numbers doesn't have an ordering of this
sort. (This is probably close to the proof you saw)

The notion of well-ordering is a set-theoretic construction, amounting
to the listing of the elements of a set one at a time (just taking a
long time, if you're going to be literal about it). The well-ordering
principle is the statement that EVERY set can be well-ordered. No
relation between the ordering and any algebraic structure, no sense of
continuity, in case your original set is somehow endowed with a
topology, just some listing of its elements in such a fashion that
every subset has a "first element" with respect to this order.
Everyone's favorite well-ordered set is the set of natural numbers

        1,2,3,4,...

The favorite well-ordered set (oops, may fail to be a set [technical
point, don't worry about it]) among mathematicians is that of the
ordinals (or ordinal numbers), because it consists of all possible
magnitudes of well-ordered sets:

     0,1,2,3,4,...,w,w+1,w+2,...2w,2w+1,...,3w,3w+1,...,w^2,..,
                   ^                        w^3,...,w^w,...,W,... etc.
                  lower-case omega        upper-case Omega^

The only problem here is that this is big, so big that there is
disagreement as to how big it is. However, it is agreed that this
"set" is well-ordered (using its natural ordering).

An example of a set with a natural ordering which isn't a well-
ordering is the set of real numbers, using the standard ordering  <.
To see this, consider the open interval (0,1). This subset has no
least element (you may say 0 is its smallest element, but then note
that 0 isn't even in the set). The reals can (by the Well-Ordering
Principle) be well-ordered, however. How? Take a 1:1 correspondence of
the set of reals with an initial segment of the ordinals, and use the
well-ordering of that segment to define the ordering on the reals.
Simple? Sure. Satisfactory? Depends on how skeptical you are about
infinite sets.

-- Dale

[ Editor: the bracketed "[technical point, don't worry about it]" is
Dale's own.  Strictly, "every subset" in the definition above means
every nonempty subset. ]