The question is about well-ordering $\mathbb R$ in ZF. Without the Axiom of Choice (AC) there exists a set that is not well-ordered. This could occur two ways: a) there are models of ZF in which $\mathbb R$ is well ordered and some other set isn't; or b) there are models in which $\mathbb R$ itself can not be well-ordered. Both possibilities are consistent with ZF.
Do I have this right?
If so then I am confused about the following argument that $\mathbb R$ must be well-ordered even in the absence of choice.
Let $\mathfrak c$ be the cardinality of $\mathbb R$. It's easy to show that $\mathfrak c = 2^{\aleph_0}$.
Now (this is the part that's got me confused) in set theory cardinals are defined as certain ordinals. Therefore $\mathfrak c$ is some ordinal; which immediately induces a well-order on $\mathbb R$.
What is wrong with this argument? I can imagine a couple of scenarios but without much confidence.
$\mathfrak c$ is a cardinal but it is possible that it's not well-ordered. But if it's a cardinal that's not an ordinal, then what is it? This scenario seems unlikely.
Or perhaps $\mathfrak c$ is not always well-defined. When people say, "Let $\mathfrak c$ be the cardinality of $\mathbb R$," they typically neglect to prove that such a thing exists. Perhaps there are models of ZF in which $\mathbb R$ has no well-defined cardinality at all. This seems like the only sensible way out, in which case there are a lot of misleading expositions out there.
I found several Stackexchange questions about properties of well-orderings of $\mathbb R$ under AC; but nothing about this particular question. According to this article the reals might not be well-ordered in Solovay's model in which all sets of reals are Lebesgue-measurable; but this requires an inaccessible cardinal. What happens if we take the inaccessible away?
In general what is the correct way to understand well-ordering the reals in ZF?
Tl;dr: the cardinality of a set need not be a cardinal. Yes, this is terrible terminology.
The former possibility is what happens if $\mathbb{R}$ is not well-orderable. $\mathfrak{c}$ is always well-defined; it's just that, in the absence of choice, it need not be an ordinal.
Note that this means that we need to be careful with our terminology when working in just $ZF$: we need to distinguish between cardinals in the sense of initial ordinals (equivalently, cardinalities of well-orderable sets) and cardinalities in general.
For that matter, we need to be a little careful how we define a possibly-non-well-orderable cardinality! We use the notion of rank. The rank of a set is defined in terms of the cumulative hierarchy. We define $V_\theta$, for every ordinal $\theta$, inductively as follows:
$V_0=\emptyset$,
$V_\lambda=\bigcup_{\beta<\lambda} V_\beta$ for $\lambda$ limit, and
$V_{\alpha+1}=\mathcal{P}(V_\alpha)$.
Then by the Axiom of Foundation, we can show that for all $A$ there is some $\theta$ such that $A\in V_\theta$. The least such $\theta$ is the rank of $A$. Moreover, each $V_\theta$ is a set.
That is, we take a Fregean approach - a cardinality is a set of all sets of a given size - but use the notion of rank to make the result a set, instead of a proper class (this is called Scott's trick).
Note that, however, this clashes slightly with the definition of the cardinality of a well-orderable set; but this doesn't really cause any problems.
By the way, I think Solovay's model is a bit of a red herring, here. The most straightforward way to break the well-orderability of $\mathbb{R}$ is Cohen's original model; Jech's big book has a good exposition of it. This does require forcing, though, so is a serious chunk of work to get through.