The theorem $157$ (p. no $275$) claims:
If $\alpha$ is an ordinal number and $x\subset \alpha$, then $\mbox{card } x \leq \alpha$.
To prove that he mentions two previous result:
Theorem $94$. If $X$ is a well-ordered class, $Y\subset X$ is a section (a segment) and $f:Y\rightarrow X$ is an order-preserving map, then for each $y \in Y$ is false that $f(y)<y$.
Theorem $99$. If $\mathcal R$ well-orders $X$ and $\mathcal S$ well-orders $Y$, then there is a function $f$ which is $\mathcal R$-$\mathcal S$ order preserving in $X$ and $Y$ and such that either $\mbox{dom }f=X$ or $\mbox{im }f = Y$.
Moreover, there one important result that I think can help us:
If $X,Y$ are two well-ordered classes and $f:X\rightarrow Y$ is an order-preserving function, then $f$ is injective and $f^{-1}$ is order-preserving.
Now, I write the proof literally as Kelleys does it:
''By theorem $99$ there is a $1-1$ function $f$ which is $\mathcal E$- (the oreder relation for ordinals) order preserving and such that $\mbox{dom }f = x$ or $\mbox{im }f=\mathcal O$ (the class of all ordinals). Since $X$ is a set and $\mathcal O$ is not, then $\mbox{dom }f = x$. By Theorem $94$, $f(u)\leq \alpha$ for all $u$ in $x$, and consequently $x$ is equivalent to an ordinal less than or equal to $\alpha$''.
EDIT:
hanks to some comments of @CarlMummert I have only one question:
I can't see how he uses Theorem $94$. It is true that $\alpha$ and $\mbox{im }f$ are sections, but I can't see how from that fact he claims that $f(u)\in\alpha$ for all $u\in x$.
I have also tried to prove the theorem on my own. Here is the proof. I would like you to ckeck it, because a have a question:
Using theorem 99 it is clear that $\mbox{dom }f=x$. By the same theorem $\mbox{im }f\subset\mathcal O$ is a section (initial sectment) and thus, by a previous result, an ordinal number. Call it $\beta$. What I want to show know is that $\beta\leq \alpha$, because then we would have an equivalence $\tilde f:x\rightarrow \mbox{im }f$ between $x$ and an ordinal $\beta$ less or equal than $\alpha$. Hence by the definition of cardinal number the theorem follows.
Because $f(\gamma)$ is an ordinal for every $\gamma\in x$, by trichotomity of $\in$:
$$ f(\gamma)<\alpha \quad \text{or} \quad \alpha<f(\gamma) \quad \text{or} \quad f(\gamma)=\alpha .$$
Suppose there is a $\gamma\in x$ such that $\alpha\in f(\gamma)$. By the above claim, I know $f^{-1}$ is order-preserving, so I would have
$$ f^{-1}(\alpha)\in\gamma \in\alpha. $$
But $\beta$ was a section of $\mathcal O$, so by theorem 94 the above is impossible, which means there is no such a $\gamma$. The same happens if I suppose $\gamma\in x$ such that $f(\gamma)=\alpha$. Hence $\mbox{im }f \subset\alpha$ and also $\mbox{im }f \in\alpha$, as I wanted.
But the problem of my proof is that $\alpha$ is supposed to be in $\mbox{im }f$ and that might not happen, might it? So... is my proof correct? If not, why can solve the problem of $\alpha$?
Thanks
I think my proof is right. If $\alpha\notin \mbox{im }f$ we have already finished. Otherwise, must supposed $\alpha\in\mbox{im }f$.