$A'$ is the set of accumulation points of $A$.
I first proved that $A'\neq\emptyset$, after I tried construct a Cauchy's sequence to show anything, but it's not so useful...
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1What is $A'$ supposed to represent? – Abhijeet Vats Aug 18 '22 at 23:13
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1acumulation points of $A$ – Kempa Aug 18 '22 at 23:19
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3Use the fact that isolated points of $A$ form an at most countable set. – geetha290krm Aug 18 '22 at 23:23
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Indeed! How $\mathbb{R}$ is separable then subset so is, I forget. Thanks. – Kempa Aug 18 '22 at 23:29
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1@Kempa A little warning: in general, a subspace of a separable space need not be separable. However, $A$ must indeed be separable: $\mathbb{R}$ is second-countable, a subspace of a second-countable space is itself second-countable, and every second-countable space is separable. – Samuel M. A. Luque Aug 18 '22 at 23:40
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Where I can read more about this? @SamuelM.A.Luque – Kempa Aug 20 '22 at 17:37
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@Kempa Any basic general topology text would work: the famous Topology by Munkres, for example. – Samuel M. A. Luque Aug 21 '22 at 09:17
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If $x \in A\setminus A'$, then there must be some $\epsilon=\epsilon(x)>0$ such that $(x,x+\epsilon) \cap A =\emptyset$. We can select some rational number $q(x) \in \mathbb Q \cap (x,x+\epsilon)$. The mapping $x \mapsto q(x)$ is injective, so $A\setminus A'$ is countable.
Yuval Peres
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