Without axiom of choice, it is not generally true that the class of all cardinals (in this question we consider Scott cardinal rather than cardinals as ordinals) is not well-founded under the ordinary cardinality comparison. However, we also know that it is fine to assume the well-foundedness of the class of all cardinals.
Under ZF with this assumption, we can prove that every infinite set is Dedekind-infinite as follows: for infinite set $X$, consider the collection of cardinals $$\mathcal{A} = \{|A| : A\subseteq X \text{ and $A$ is infinite}\}.$$ From assumption, $\mathcal{A}$ is well-founded. If $|B|$ is a minimal element, then $B$ should be Dedekind-infinite, since $|B|-1 = |B|$. (where $|B|-1$ is a cardinality of the set $B$ except one element in $B$.)
I wonder we can prove a stronger result; for example, the axiom of choice follows from that the class of cardinals is well-founded? I would appreciate your answer.
This is an open problem.
It was shown that for every $\kappa$, $\sf DC_\kappa$ cannot prove that the cardinals are well-founded. While not enough to conclude the principle is equivalent to the axiom of choice ($\sf BPI$ does not follow from $\sf DC_\kappa$ either), it is worth remarking that we really don't know much about this principle.
A very recent paper gave a nice survey of this problem and related results:
Let me finish by stating that generally speaking the structure of the cardinals is a bit of a wild beast when it comes to the axiom of choice. We don't have good techniques to control it very well in order to produce separating models for much awaited-results (e.g. the Partition Principle is a statement about the structure of the cardinals). So we mainly know how to violate things wildly (e.g. embed partial orders into the cardinals of a model), but not how to fine tune this in order to produce nice results.