Sets of Cardinals without choice

Solution 1:

If every set of cardinal numbers is well-ordered by cardinality, then in particular any two cardinals are comparable by cardinality (i.e., trichotomy holds).

To see that trichotomy implies the well-ordering theorem: Let $X$ be any set that we want to well-order, and let $\alpha$ be its Hartogs number, which is the first ordinal that cannot be injected into $X$. (The existence of such an ordinal can be proved without Choice; it is the union of the ordinals that represent the order types of relations that well-order subsets of $X$). Since by assumption $|\alpha|\le|X|$ is not true, trichotomy tells us that $|X|<|\alpha|$, so there exists an injection $X\to\alpha$, and therefore $X$ can be well-ordered.