Fact 1: An ordinal $\alpha$ is a cardinal iff $\xi < \alpha \Rightarrow |\xi|<|\alpha|$. (This is the definition of cardinals; see, e.g., 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: Now, 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.)