The binary tree and a cofinal branch

The predecessors of $s:n\to 2$ are exactly $s\upharpoonright k$ for $k<n$, ordered like $n$. Thus its height is $n$. There is one empty string at level zero; appending either $0$ or $1$ to each length-$n$ string gives all length-$(n+1)$ strings without repetition. Induction gives $\mathrm{\mid }{T}_n\mathrm{\mid }=2^n$, including $2^0=1$.

Put $z_n(k)=0$ for $k<n$. Then $z_n=z_{n+1}\upharpoonright n$, so $B=\{z_n:n<\omega \}$ is an infinite chain. A string $s$ of length $n$ comparable with every $z_k$ must equal $z_n$, since comparable strings of equal length coincide. Thus $B$ is maximal and is a cofinal branch.

If $S\subseteq 2^{<\omega }$ is infinite and prefix closed, it has finite levels, each of size at most $2^n$. Its lengths cannot be bounded by $N$, since then $\mathrm{\mid }S\mathrm{\mid }\le {\sum }_{n=0}^N{2}^n=2^{N+1}-1$. Its height is therefore $\omega$, and F1 supplies an infinite branch. This last appeal inherits the ZFC assumption; the explicit branch in step 2.1 needs no choice.