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Feb 23, 2022 at 6:56 comment added Alessandro Codenotti @bof that's a nice way to write it, thanks! I struggled to find a concise way to say what I wanted
Feb 23, 2022 at 6:52 vote accept Dominic van der Zypen
Feb 23, 2022 at 0:07 comment added bof More concisely, if an infinite $T_1$ space is coloured with finitely many colours, consider an infinite open set with a minimum number of colours, and delete a "uniquely coloured" point to get a contradiction.
Feb 22, 2022 at 22:20 comment added Alessandro Codenotti Really all I'm saying without making it harder than it has to be is that there is a point $x_1$ with a unique colour in $\Bbb R$, so there is $x_2$ with a unique colour in $\Bbb R\setminus\{x_1\}$, so there is $x_3$ with a unique colour in $\Bbb R\setminus\{x_1,x_2\}$ and all those points have different colours.
Feb 22, 2022 at 22:14 history edited Alessandro Codenotti CC BY-SA 4.0
added 1063 characters in body
Feb 22, 2022 at 21:56 history answered Alessandro Codenotti CC BY-SA 4.0