I am trying to understand how a normal tree with $\aleph_1$ node can fail to have a branch of cardinality $\aleph_1$.
Consider the tree of height $\omega_1$ whose nodes are $\mathbb{Q}$-valued strictly increasing $\alpha$-sequences.
This tree is normal : it has a single root; every node has a least to immediate successors; for every $\alpha < \beta < \omega_1$, if $x$ is on the $\alpha$-th level then there exist $y$ on the $\beta$-th level such that $x < y$.
This tree cannot have a branch of length $\omega_1$, for there would be an injection of $\omega_1$ into $\mathbb{Q}$.
Why does a proof in the style of König's Lemma break down for ω1? What is wrong with "Pick a node that has an uncountable number of successors as, if one doesn't exist, then you have a countable union of countable things which must (choice) be countable."?
This question is closely related to Aronszajn Trees and König's Lemma but but I am not requiering that the tree be an Aronszajn tree.
How are you going to carry the recursive construction of a branch beyond the first $\omega$ steps? Every node has uncountably many successors at the next higher limit level, so you can easily construct a strictly increasing sequence $\langle x_n:n\in\omega\rangle$ of nodes, but if $\sup_n\sup x_n=\infty$, which is certainly possible, you won’t be able to extend this sequence of nodes any higher in the tree.