I'm not sure I correctly remember what $p$-good means, but note that the mod $p$ homology of $B\mathbb{Q}$ is of course zero.

Yes, I think if you want to turn this idea into an actual proof you look at the induced map of fiber sequences $B\mathbb{Q}\to B\mathbb{Q/Z} \to K(Z,2)$ and $B\mathbb{R}\to B\mathbb{S^1}\to K(Z,2)$, and observe that the map on the fibers becomes an equivalence after $p$-completion.

Well, this tells you that $a$ goes to the permutation (on the set of triples $(k, l, m)$ with the bounds mentioned in the answer) which sends $(k, l, m) \mapsto (k, k+l, 1+(k+l)p+m)$.

@Wlod AA, I'm pretty sure that way to use edits is against the rules here. If you have something to say, try comments or answers...

Maybe the problem is that literally as written, your topology is not a topology, since the empty set also needs to be open.

"Continuity and having the fixed point $x_0$ are one and the same thing here.", that's not true, is it? For example if $x_1$ is any other point, the constant map to $x_1$ is also continuous with respect to your $x_0$-topology, since all preimages are either empty or everything. But this does not have $x_0$ as fixed point.

If I understand your definition correctly (and I'm absolutely not sure whether I do, it would help if you gave a clear definition of what you mean by fundamental ring), then it doesn't sound like the fundamental ring will be a homotopy invariant. Compare for example the interval and the point. The fundamental groupoids of two homotopy equivalent spaces are equivalent (in the sense of equivalence of categories)

Ah, I guess one can directly make this explicit: Your $\mathrm{Tor}_1$ is just the $\ell^n$-torsion subgroup of $\pi_{-1}(K)$, and the latter is divisible, so the maps in the diagram for your $\lim^1$ are all surjective. By the way, using that connectivity is detected mod $\ell$ one can also directly do the original question: For $X$ $\ell$-complete with perfect mod $\ell$-reduction, inductively build a finite complex of copies of $\mathbb{Z}_\ell$ with map to $X$, which induces an equivalence mod $\ell$. This is formally similar to how one builds minimal CW approximations etc.

Ah, I see, that's a subtlety I missed, sorry. But I think for a derived $K$-complete complex you can check connectivity mod $\ell$. To see this, assume $K/\ell$ is connective. Then $\pi_k K$ for $k<0$ are divisible. But they are also derived $\ell$-complete, and so they are zero.

That's a completely elementary exercise, you just check that a map between discrete $\ell$-complete $\mathbb{Z}_\ell$-modules is surjective if it is surjective mod $\ell$.

This shows that the mod $\ell$-reduction of homology, $H_*(X)/\ell$, maps injectively into $H_*(X/\ell)$, so if the latter is finitely generated, so is the former.

Well, for any complex $X$, you have a cofiber sequence $X\xrightarrow{\ell} X\to X/\ell$, and this induces a long exact sequence in homology.