Let $T$ be a normal Suslin tree. We define a set $S$ to be the collection of branches in $T$. Given an element $x \in T$, it has $\omega$ successors in the level above - order these in the same way as $\mathbb{Q}$. Now we construct a dense linear ordering for $S$: given two branches $a,b$ in $T$, let $\alpha$ be the smallest level where the branches differ. The two differing points $a_{\alpha}$ and $b_{\alpha}$ must lie on a successor level and are immediate succesors of a point in the level below, so we can order them as rational numbers - we say $a < b$ if $a_{\alpha} < b_{\alpha}$ and $a > b$ if $a_{\alpha} > b_{\alpha}$.
Now given an open interval $(a,b) \subseteq S$, I want to show there is some $x \in T$ such that $I_x \subseteq (a,b)$, where $I_x := \{ c \in S \mid x \in c\}$. I'm a little bit stumped here, can anybody help me out? My text (Jech) claims that $I_x$ is an interval, i'm not sure why this is true either. Any help would be appreciated.