Cardinality of all cardinalities

$C$ is not a set $-$ it is infact a proper class. If $C$ were a set, then $|C|$ would be defined. It then follows that $|C|$ would be the largest cardinality, since there is a total order between all the cardinalities, and $|C| > \kappa$ for every cardinality $\kappa$ (every cardinality is equivalent to the set of all smaller cardinalities). But $2^{|C|} > |C|$ and so there cannot be a largest cardinal.

This is "Fact 20" on page 10 of

http://math.uga.edu/~pete/settheorypart1.pdf

These are notes on infinite sets from the most "naive" perspective (e.g., one of the facts is that every infinite set has a countable subset, so experts will see that some weak form of the Axiom of Choice is being assumed without comment. But this is consistent with the way sets are used in mainstream mathematics). It is meant to be accessible to undergraduates. In particular, ordinals are not mentioned, although there are some further documents -- replace "1" in the link above with "2", "3" or "4" -- which describe such things a bit.

But I don't see why it is necessary or helpful to speak of ordinals (or universes!) to answer this question.

Added: to be clear, I wish to recast the question in the following way:

There is no set $C$ such that for every set $X$, there exists $Y \in C$ and a bijection from $X$ to $Y$.

This is easy to prove via Cantor's diagonalization and it sidesteps the "reification problem for cardinalities", i.e., we do not need to say what a cardinality of a set is, only to know when two sets have the same cardinality. I believe this is appropriate for a general mathematical audience.

The most common way to define the cardinal number $|X|$ of a set $X$ is as the least ordinal which is in bijection with $X$. Then $C$ is an unbounded class of ordinals, and any such is necessarily a proper class. Since $C$ is not a set, it does not have a cardinality.