Branches through countable normal trees of limit height

Fix surjections $e:\omega \to T$ and $d:\omega \to \delta$. They exist by countability: $\delta$ is infinite, and $T$ has nodes at arbitrarily high levels below $\delta$, so it too is infinite. Only these two witnesses are fixed. Define ${\gamma }_0=d(0)$ and ${\gamma }_{n+1}=\max \{{\gamma }_n+1,d(n+1)\}$. The successor of every ordinal below the limit $\delta$ is still below $\delta$, so recursion gives a strictly increasing sequence in $\delta$. It is cofinal since ${\gamma }_n\ge d(n)$. The nonautonomous rule is a recursion on the state $(n,{\gamma }_n)$.

Starting at $t_0=t$, let ${\beta }_n=\max \{{\gamma }_n,\mathrm{ht}(t_n)+1\}<\delta$ and take $t_{n+1}=e(k)$ for the least $k$ for which $t_n{<}_{T}e(k)$ and $\mathrm{ht}(e(k))={\beta }_n$. The candidate set is nonempty by normality. Recursion on $(n,t_n)$ supplies this sequence, and its heights are cofinal because $\mathrm{ht}(t_{n+1})\ge {\gamma }_n$. Least natural indices require no choice function.

Put $b_t=\{s\in T:\mathrm{\exists }ns{\le }_T{t}_n\}$. It contains $t$ and is a chain: if $s{\le }_T{t}_n$ and $u{\le }_T{t}_m$, both lie below $t_{\max \{n,m\}}$, so they are comparable. Its heights are cofinal by step 2.1.

If $s$ can be adjoined to $b_t$ while retaining a chain, choose $n$ with $\mathrm{ht}(t_n)>\mathrm{ht}(s)$, possible by cofinality and the limit-height hypothesis. Comparability with $t_n$ and strict increase of height force $s{<}_T{t}_n$, hence $s\in b_t$. Thus $b_t$ is maximal and is a cofinal branch.

The same fixed $e,d$ determine $b_t$ uniquely for each $t\in T$. Replacement therefore forms $\mathcal{B}=\{b_t:t\in T\}$. The sequence $n\mapsto b_{e(n)}$ is onto $\mathcal{B}$, so it is countable, and $t\in b_t$ proves that it covers $T$. The construction includes the root and every prescribed node; no last level exists at the limit height.