I'm a bit confused on Jech's proof of Lemma 9.13. I will not give the definitions of "normal Suslin tree" or "Suslin tree" here.
He claims that if $T$ is a Suslin tree, and $T_1 = \{x\in T:\; T_x$ is uncountable $\}$, then if $x\in T_1$ and $\alpha>o(x)$, then $|T_y|=\aleph_1$ for some $y>x$ at level $\alpha$. My first question is why this is the case.
Secondly, why does this imply that $T_1$ has property (v) like he says right afterwards? (v) says that for each $x\in T_1$ there is some $y>x$ at each higher level less than $\omega_1$.
Given some level $\alpha>o(x),$ we have $$T_x = \{y\in T_x: o(y) < \alpha\}\cup\bigcup\{T_y: o(y)=\alpha, y> x\}.$$ So, since the levels are all countable, if $T_y$ were countable for all $y>x$ at level $\alpha,$ then $T_x$ would be a countable union of countable sets.
As for the second part, this is perhaps confusing because we have just shown something much, much stronger than what Jech has asked us to conclude. Of course if $|T_y| = \aleph_1$ for some $y>x$ at level $\alpha,$ then there is some $y>x$ at level $\alpha$!