hrmm. construct for any ordinal α S_α

S_0=S
S_{α+1} is S_α minus all its isolated points

S_λ is the intersection of S_α α<λ

S_α is always closed, nonempty, and infinite

S_α is also eventually constant if for no other reason than that eventually you run out of points. call the resulting set T

pretty sure because T is perfect and this is a nice enough topological space T has cardinality continuum

i think this reasoning still works if we're talking tiles where there's only a finite number of ways to glue 2 tiles together sensibly
you don't quite get the group action but you do still get a compact topology on all the ways to build a tesselation around some base tile

and one point in this space always gives you an infinite number because aperiodic

you feel like there has to be a one-line proof citing some theorem in model theory or something

i feel like you can repeat the argument in full generality with some bespoke metric on the space of all tilings where the tiles are allowed to slide around but that seems like effort

(in actuality S_α will become constant while α is still countable
because our basis is countable
and the set of things B_α in our basis which are a subset of the open set S\S_ α always gets bigger as α goes up
and as S\S_α is the union of all things in B_α...)

https://twitter.com/CihanPostsThms/status/1644700418583273473
here's a result you get with very similar argument lol
---
RT @CihanPostsThms
The following is a theorem of ZFC (in particular CH is not assumed):

Let G be a group of cardinality ≤ |ℕ|. Then the cardinality of the set of subgroups of G is
• either ≤ |ℕ|,
• or equal to |ℝ|.
https://twitter.com/CihanPostsThms/status/1644700418583273473