Is infinitary Böhm-reduction wrt. root-active terms for $\lambda$-calculus transitive?
I think you'll find that the exact same proof of Lemma 3 in [1] (the proof itself appears in [2]) concerning $\rightarrow^\infty_\beta$ also holds for $\rightarrow^\infty_{\beta\bot}$: indeed, they are defined in the same way from $\rightarrow^*_\beta$ and $\rightarrow^*_{\beta\bot}$ respectively, which are transitive by definition!

The lemma holds for an arbitrary reflexive-transitive congruence relation in place of $\rightarrow^*_{\beta\bot}$, as you can verify in [2] without needing to go into the detail of the coinductive proofs. In particular, the proofs of lemmas 4.1-5 in [2] are unchanged (with extra induction cases in 4.4).