The proof: Let $\alpha= \sup X$. If $f$ is a bijective mapping of $\alpha$ onto some $\beta < \alpha$, let $\kappa \in X$ be such that $\beta < \kappa \le \alpha$. Then $|\kappa|=|\{f(\xi): \xi < \kappa\}| \le \beta$, a contradiction. Thus, $\alpha$ is a cardinal.
*An ordinal $\alpha$ is a cardinal if $|\alpha|\neq |\beta|$ for all $\beta < \alpha$.
Here's how far I think I understand the logic of the proof; first we assume $\alpha$ isn't a cardinal. This means there would be a $\beta$ less than $\alpha$ which is equinumerous to $\alpha$. We then suppose another ordinal $\kappa$ greater than $\beta$ but that injects into $\alpha$. What I don't get is how do we know that $|\kappa|=|\{f(\xi): \xi < \kappa\}|$, not to mention how $|\{f(\xi): \xi < \kappa\}| \le \beta$. If all of that stands true, then I get that we arrive at the contradiction; $\beta < \kappa$ means $\beta \in \kappa$ and that implies there can't be an injection from $\kappa$ into $\beta$ ($\kappa \le \beta$).