I'm reading Jech's book "Set Theory" precisely in the part about Suslin trees. On Lemma 9.13 he proves that if there exists a Suslin tree, then there exists a normal Suslin tree.
If $T$ is a normal tree of height $\alpha$, then, in particular, it satisfies that
For any $x\in T$ and $height(x)<\beta<\alpha$, there exists $y>x$ such that $height(y)=\beta$.
But it just seems weird for me, because this property seems to imply, by transfinite induction, the existence of a chain with the same height of $T$. But if this is true, then it should be no normal Suslin trees, since every Suslin tree has height $\omega_1$ and no uncountable chains. What am I missing here?
You are missing the limit stages. Sure, if a branch is successor-length (i.e. has a maximum element), you can keep extending it level by level for $\omega$ more steps, but how do you know that it keeps going after that? For every node on the branch, there is some node above it at the next limit level, but that doesn't mean you can say there's a node above the whole branch.
For a more concrete example of a normal tree with height $\omega_1$ and no branch of length $\omega_1$, consider the tree of functions $f$ such that $f$ is injective, dom($f$) is a countable ordinal, and ran($f$) is a subset of $\omega$ whose complement is infinite.