Thank you for this detailed answer! I will accept it as it settles that my claim is unprovable.

@MichalAdamaszek Would you be able to write up the full answer? I am having trouble visualising the counterexample. Thanks for your comment!

@AchimKrause Yes, sorry, to be honest I wasn't super careful there but thanks for formalising it. I hope I'm not barking up the wrong tree with this $A\cap B$ business. But I'll keep trying to prove that $H_{k-1}(A\cap B)$ is $0$ and update if I get anything.

Ah, you're right---I picked the worst possible letter for $X_n$. I'm not going to change it now, but I will add a note above.

Because there are now far fewer vertices to work with, it almost seems like one can work by induction. There is a simplex consisting of the multiples of every prime, and whenever a new vertex $n$ is added, it becomes a vertex of intersection among all the simplices corresponding to primes that divide it. It feels like this should never cause any empty cycles to appear, but I'm still thinking about how to prove this rigorously.

Thanks for the answer anyway! I have thought about this easier simplicial complex as well, and thought it was fun that Bertrand's postulate plays a role here.

Wow, thanks! I do think this will be easier to work with than the original complex. I think it is okay to keep $1$ in the index set, since by the definition above, the set $\{1\}$ is a maximal coprime-free set (and there are already other isolated vertices for every prime $p$ in $(n/2,n]$, so adding $1$ alongside these primes shouldn't mess things up too much). Besides these primes and $1$, the rest of $\Delta_n$ should be connected.

@HenrikRüping Ah, yes, I think these two complexes are equivalent homology-wise, but did not consider using the chain version for this problem---I will think about it.

@AchimKrause As HenrikRüping said, the way $\Delta_n$ is defined does not allow one to do this. If we wanted one of the sets to contain $35$, we would also have to include smaller values such as $30$.so as not to contradict maximality.

Thanks for this answer and the really nice construction! I (perhaps rather crudely) bounded the density of $E$ from below by $1/2$, so there is a set of density at least $1/8$ without any factorial differences.