EDIT: this question is due to a confusion. Noah answered it appropriately. However, a solution of the more general theorem has been added as an answer. In particular, it is a proof of the theorem that I thought was proven in Drake's reference.
I was reading Set Theory, An introduction to Large Cardinals, by Frank Drake (1974), and I found the following proof of König's Lemma for arbitrarily large trees (in particular, trees on any cardinal with finite levels):
For clarification: "finitary tree" is defined there as a tree with finite elements in the first level and such that any element has finitely many successors and $T_\alpha$ denotes the subset of the tree whose elements are below the $\alpha$-th level. In addition, the concept of "infinite tree" is not defined explicitly, but it is interpreted by the context (for example, the last sentence of the proof), that they refer to trees on an arbitrary infinite cardinal.
My first problem is that I don't see how this proof (from the second paragraph onwards) fails for trees with non-finite levels. For example, if $|T|=\omega_1$ and the levels are countable, I think we could still say "pick $t_0\in T_1$ such that $t_0$ has $\omega_1$-many successors in $T$, there must be one since $|T|=\omega_1$ and $|T_1|=\omega$". And similarly for the $n$-th step. Therefore, this would be against Aronszajn's theorem (this is, there exist an Aronszajn tree).
Secondly, I thought that the critical step for this prove was the limit step, but only the successor step is taken into account during the proof…
Regarding the two previous issues, how does this proof work? Thanks in advance!!
Best Answer
The point is that the proof is only trying to build an infinite path, not a path of length $\omega_1$. If you want a path of the latter type you need to figure out how to "continue past $\omega$," but if you want a path of the former type you just need to figure out how to keep going for each natural number.
In particular (and re: your proposed counterexample), the following is true: if $T$ is an uncountable tree with each level countable, then $T$ has an infinite path. However, this doesn't contradict the existence of Aronszajn trees, since the requirement for the latter is that they have no uncountable (not just infinite) path. An Aronszajn tree $T$ does have lots of infinite paths, it's just that all the infinite paths through such a $T$ are "much shorter than" $T$ itself.