This is a question out of idle curiosity.
There’s a famous quip that “the axiom of choice is obviously true, the well-ordering principle is obviously false, and who can tell about Zorn’s lemma?”
Let us imagine a hypothetical typical mathematician “Bob”, who thinks that the well-ordering principle is obviously false. Certainly Bob would still agree that the well-ordering principle holds for countable sets, since $mathbf N$ can be well-ordered and then also any set in bijection with $mathbf N$. Now if one believed that there exists a well-ordered set of cardinality continuum, then one would be forced to say that it is similarly obvious that $mathbf R$ can be well-ordered, by the same argument. So Bob does not believe that a well-ordered set of cardinality $mathfrak c$ exists, and perhaps even more generally does not believe in the existence of any uncountable well-ordered set.
And there is indeed something unintuitive about uncountable well-ordered sets (at least to me, and to Bob). But we do not need the axiom of choice to exhibit them. ZF proves Hartog’s lemma:
For any set $X$ there exists a well-ordered set $Y$, such that there does not exist an injection $Y hookrightarrow X$.
In the presence of choice the non-existence of an injection $Y hookrightarrow X$ implies the existence of an injection $X hookrightarrow Y$ and hence a well-ordering of $X$. Now our hypothetical mathematician Bob probably has no issue with the fact that cardinals are totally ordered — at least I’ve never heard anyone express this as an unintuitive consequence of choice — so if Bob does not believe in the well-ordering principle then he probably does not believe in Hartog’s lemma, either, meaning that the axiom of choice was not at all the thing that Bob was skeptical about.
My question is whether there exists an alternative axiomatic set theory in which Bob’s intuition is correct, i.e. in which Hartog’s lemma fails (and preferably the axiom of choice is true), and in which all ordinals are countable. In Bob’s dream world the axiom of choice is true, say in the form that a product of nonempty sets is nonempty, but transfinite recursion has instead been crippled somehow to the point where choice can not be used to prove unintuitive statements like Banach-Tarski and the existence of a well-ordering of $mathbf R$.
In ZF the collection of all countable ordinals form a set, which is then the smallest uncountable ordinal $omega_1$. So in Bob’s set theory this can not be a set. My guess would be that Bob does believe that the collection of all well-orderings of subsets of $mathbf N$ is a set. This uses power set and separation. To get from here to having the class of countable ordinals being a set uses only the axiom of replacement, so my guess is that Bob wants to throw out replacement as being too powerful. But I’m not sure how much of set theory and everyday mathematics breaks without replacement.