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Sep 21, 2014 at 11:45 vote accept Fedor Petrov
Sep 19, 2014 at 16:14 comment added Ramiro de la Vega @LeeMosher: It means that $I$ is a subset of $\omega_1$ (the one inside the square brackets) of size $\omega_1$ (the one in the superscript). In some contexts you may want to change "size" by "order type" but it doesn´t make a difference here.
Sep 19, 2014 at 15:59 comment added Lee Mosher What does the notation $I \in [\omega_1]^{\omega_1}$ mean? That's the one place where this topologist gets lost.
Sep 19, 2014 at 12:43 comment added Emil Jeřábek (Nice argument, by the way.)
Sep 19, 2014 at 11:40 comment added Emil Jeřábek Oh, I see. So this is not meant as a standalone answer, but as a replacement of the last part of your (Joel’s) proof? @Ashutosh, I think you should make this clear in the answer.
Sep 19, 2014 at 11:32 comment added Joel David Hamkins Emil, in the final case of my answer, I get uncountably many points of a fixed distance apart, and such that they are isolated as well. So the set Ashutosh constructs is open in the whole space, not just in the subspace. So this argument exactly handles the case where I had used CH.
Sep 19, 2014 at 11:12 comment added Emil Jeřábek I don’t understand how this answers the question. If $X$ is a counterexample to the OP, it does contain an uncountable subspace $Y$ whose points are at least a fixed distance apart. Now, any open set $U$ is a countable union of balls in $X$, however their centres may be outside $Y$, hence it doesn’t follow that $U\cap Y$ is a countable union of balls in $Y$.
Sep 19, 2014 at 9:35 comment added Joel David Hamkins Great! This seems to get rid of the need for any CH assumption in the final case of my argument.
Sep 19, 2014 at 9:29 comment added Bill Johnson Nice conclusion.
Sep 19, 2014 at 2:37 history answered Ashutosh CC BY-SA 3.0