I am reading some set theory right now and started to read about the Axiom of Choice and that ZF is consistent iff ZFC is consistent. So, since the axiom of choice causes the cardinalities of sets to have a similar structure as the natural numbers (simplified) I wanted to ask if it's possible to add an axiom to ZF so that the cardinalities have a "dense" structure (like the real numbers) meaning is ZF $\cup$ D consistent iff ZF is consistent?
where D:=for any sets $M,N$ s.t $M,N$ is infinite and $|M| < |N|$ there exists a set $B$ s.t $|M|<|B|<|N|$
No. If $N$ is well-orderable, then every subset of $N$ is also well-orderable (just restrict the ordering), so the usual theory of well-ordered cardinalities applies to all subsets of $N$. In particular, for instance, if $N=\omega_1$ and $M=\omega$ then no such $B$ can exist, since every infinite subset of $N$ has cardinality $\aleph_0$ or $\aleph_1$.