Countable levels do not suffice for König’s lemma

The predecessors of a sequence $s$ of length $n$ are exactly its restrictions to lengths $0,1,\dots ,n-1$, in that order. Thus the proper initial-segment relation is transitive and irreflexive, its predecessor orders are finite well-orders, and ${\mathrm{ht}}_T(s)=n$ by F1. The empty sequence is the unique root.

The level $T_0$ is a singleton. For each fixed $n$, the set ${\omega }^n$ of length-$n$ sequences is countable: start with the singleton ${\omega }^0$ and iterate F2 using ${\omega }^{n+1}\cong {\omega }^n\times \omega$. Since $T_n\subseteq {\omega }^n$, F3 makes every level countable. For $n>0$ the explicit sequence $(n-1,n-2,\dots ,0)$ has length $n$ and lies in $T_n$. Therefore all finite heights occur and the height is exactly $\omega$. The first level contains every $(m)$ for $m\in \omega$, so it is infinite.

If $B$ were a cofinal branch, its sequences would be nested and have unbounded lengths by F1. Their union would therefore be a function $b:\omega \to \omega$. For each $n$, take a node in $B$ of length at least $n+2$; its strict decrease gives $b(n+1)<b(n)$. The nonempty range of $b$ has a least element $b(k)$ by F4, but $b(k+1)$ is a smaller element of the same range, a contradiction. Thus the displayed countable-level tree has no cofinal branch. Its infinitely branching root and infinite first level explain precisely the failure of the finite-level hypothesis.