For any two sets $A,B$ , $|A|\leq|B|$ or $|B|\leq|A|$ Let $A,B$ be any two sets. I really think that the statement $|A|\leq|B|$ or $|B|\leq|A|$ is true. Formally:

$$\forall A\forall B[\,|A|\leq|B| \lor\ |B|\leq|A|\,]$$

If this statement is true, what is the proof ?

This is true in $ZFC$ because of Zermelo's Well-Ordering Theorem; given two sets $A,B$, since they are well-orderable there exist alephs $\aleph_{\alpha},\aleph_{\beta}$ with $|A|=\aleph_{\alpha}$ and $|B|=\aleph_{\beta}$, since alephs are comparable, the cardinalities of $A$ and $B$ are comparable.
Furthermore, this is equivalent to the axiom of choice:
Now suppose the cardinalities of any two sets are comparable, let us prove Zermelo's Well-Ordering Theorem from this. Let $X$ be a set, then there must be an ordinal $\alpha$ such that $|X|\leq |\alpha|$, for otherwise the set of all ordinals equipotent to a subset of $X$ would contain the proper class of all ordinals; for all ordinals $\alpha$, $|\alpha|\leq |X|$ by the comparability condition, a contradiction. Hence $|X|\leq |\alpha|$ for some ordinal $\alpha$, and thus $X$ is well-orderable.
Hence the condition of comparability is equivalent to Zermelo's Well-Ordering Theorem, and thus it is equivalent to the axiom of choice.
