A relation is a subset of the Cartesian product of two sets, if "have the same cardinality", denoted as $R$, is a relation, then there must exist set $A, B$, such that $R \subset A \times B$. What are $A, B$ then? They cannot be "set of all sets", because there is no such set according to axiom of regularity (ZFC set theory).
Did I miss something?
Thanks.
You are right, this is not an equivalence relation if you require that the domain be "all sets". However, if you restrict yourself to, say, all subsets of a given set $X$, then this is an equivalence relation now. (And, by picking an appropriate $X$, we can always ensure to include in the domain of the relation whatever sets we may actually be interested in.)
It is common, however, not to worry about this when arguing about cardinalities. After all, it is for sure true that for any sets $A,B,C$, we have that $A$ has the same size as $A$, if $A$ has the same size as $B$ then $B$ has the same size as $A$, and if $A$ has the same size as $B$ and $B$ has the same size as $C$ then $A$ has the same size as $C$. This is what matters when actually working with cardinalities, rather than whether something or other satisfies the formal definition of equivalence relation.
That being said, there is one reason why we may want this to be an equivalence relation, namely so that we can take equivalence classes and work with them. If we just work with the class of all sets, then the equivalence classes would be proper classes (that is, they would be "too big" to be sets), and this may lead to issues when formalizing what one is doing. In axiomatic set theory, the standard solution to this problem is provided by the axiom of foundation. This assigns to each set a rank (formally, an ordinal), which allows us, via a little detour, to define equivalence classes as sets. Another solution is provided by the axiom of choice, that allows us to pick canonical representatives from each equivalence class (certain ordinals).
