Question: Let $\alpha$ be an ordinal number. How can I show that every subset $x$ of $\alpha$ has either the same cardinality as $\alpha$ or there exists an cardinal number $\beta$ such that $|x|=\beta < |\alpha|$?

Answer: 
First prove that if $x\subseteq\alpha$, then there is an ordinal $\beta\leq\alpha$ such that $x$ is order isomorphic to $\beta$ (that is to say, $\beta$ is the order type of $x$).

Then you only need to prove that if $\beta\leq\alpha$, then the cardinality of $\beta$ is at most that of $\alpha$. This should be a fairly easy exercise in applying definitions.
Let $\alpha^+$ denote the [Hartogs number](https://en.wikipedia.org/wiki/Hartogs_number) of $\alpha$, i.e. the least ordinal of greater cardinality than $\alpha$. This concept is well-defined for any set. The Hartogs number of any $x\subseteq\alpha$ is at most $\alpha^+$. If it's exactly $\alpha^+$, $x$ bijects with $\alpha$; if not, $x$ bijects with some lesser ordinal.