I'm having a hard time following this proof. Here $\aleph(\alpha)$ is the cardinality of $Z(\alpha)$, the set of all ordinals $\gamma$ such that $|\gamma|\leq\alpha$. Also, $\aleph_1=\aleph(\aleph_0)$ and so on.
Suppose $\kappa\lt\aleph_\alpha$. Then there exists $B\subset Z(\alpha)$ with $|B|=\kappa$. Let $\gamma$ be the ordinal of $B$. Then $\gamma\leq Z(\alpha)$. But $\gamma\neq Z(\alpha)$ since then $|\gamma|=\kappa<\aleph_\alpha=|Z(\alpha)|$. So $\gamma<Z(\alpha)$, so $|\gamma|\leq\alpha$.
My one question is, why does "Let $\gamma$ be the ordinal of $B$. Then $\gamma\leq Z(\alpha)$" follow? It seems like it's almost assuming the result. All I conclude from $\gamma$ being the ordinal of $B$ is that $|\gamma|=k$, since $B$ and $\gamma$ are isomorphic, right?