Question regarding trees Let $(T,<)$ be tree of height $\aleph_2$ in which every level $L_\alpha$ is countable. A proof I am reading claims that for every $t\in L_{\omega_1}$ there is a $s<t$ which extends uniqueley to level $\omega_1$, that is with $\{t'\in T \mid s\le t'\}\cap L_{\omega_1}=\{t\}$. But I don't see why this should be true, couldn't the tree up to level $\omega_1$ just look like $\omega_1$ with the usual ordering together with two different elements $a,b$ on top?

Yes, you are right. 
Some people, however, use the word tree to mean that it satisfies a normality condition, by which every branch up to a limit level has a unique limit node. Your counterexample tree is not normal in this sense. 
Meanwhile, perhaps the theorem you are reading is the theorem asserting that every tall narrow tree of the kind you cite amounts to countably many branches, with dead parts branching off. This result will be true for tall narrow trees generally, not just normal trees. To see this, one can insert imaginary limit levels with unique nodes on top of the cofinal branches to make a normal tree, and then run the argument with these new imaginary nodes. 
