The supremum of any set of cardinals (considered as a set of ordinals) is again a cardinal. An ordinal $\alpha$ is a cardinal iff no $\xi < \alpha$ is equivalent to $\alpha$. Now, let $A$ be any set of cardinals and $\sup(A)=\alpha$, then for $\xi < \alpha$ there is a $\beta \in A$ such that $\xi < \beta$.  As such $|\xi|$ is smaller than that of $\beta$, and thus it is smaller than that of $\alpha$.
Does this sound correct/in the right direction? If so, could someone help better explain the first line?
 A: Fact 1: An ordinal $\alpha$ is a cardinal iff $\forall \xi:\ \xi < \alpha \Rightarrow |\xi|<|\alpha|$.
Proof: Easily by Proposition 6.3 of https://web.archive.org/web/20170329065559/http://euclid.colorado.edu/~monkd/m6730/gradsets06.pdf
Here $|\xi|$ denotes the cardinality of $\xi$.
Fact 2: The supremum of a set of ordinals is a well-defined ordinal, namely $\sup A:=\cup A$. See, e.g., Is the supremum of an ordinal the next ordinal?
Theorem The supremum of any set of cardinals is a cardinal.
Proof: Let $A$ be any set of cardinals and $\sup(A)=\alpha$. If $\max A$ exists, then $\alpha=\max A\in A$, hence then $\alpha$ is a cardinal.
If $\not\exists\max A$, then for any $\xi < \alpha$ there is a $\beta \in A$ such that $\xi < \beta$. But then $|\xi|<|\beta|\le|\alpha|$, by Fact 1.*
As $\xi<\alpha$ was arbitrary, $\alpha$ is a cardinal, by Fact 1, QED.
++) In fact here $|\beta|<|\alpha|$, but we don't need that fact.
(Proof: as otherwise for any $\gamma\in A$ we would have $|\gamma|\le|\alpha|=|\beta|$, hence $\gamma\le\beta$, as both are cardinals,
so $\beta=\max A$, which was ruled out.)
(This is essentially the proof given above by others; I just wanted to make it clearer.)
