Consider the following questions, how do I show countability of set $A$ and $B$?
(a) A subset $A$ of $\mathbb{R}$ has the property that, given $\varepsilon > 0$ and $x \in \mathbb{R}$, there exist $a,b \in \mathbb{R}$ with $a \in A$ and $b \not \in A$, such that $|x-a|<\varepsilon$ and $|x-b|<\varepsilon$. Can $A$ be countable? Can $A$ be uncountable?
(b) A subset $B$ of $\mathbb{R}$ has the property that, for every $b \in B$, there exists $\varepsilon > 0$ such that for every $x \in \mathbb{R}$, $0<|b-x|<\varepsilon$ implies $x \not \in B$. Is $B$ countable?
Questions come from Cambridge Mathematical Tripos
a. $A=\mathbb{Q}, \mathbb{Q}^c = \mathbb{R}/\mathbb{Q}$ work equally well, so $A$ can be either. Notice that the statement is invariant upon taking complements.
b. Suppose $B$ is uncountable. By the uncountable pigeonhole principle, some interval of the form $[n, n+1], n \in \mathbb{Z}$ contains uncountably many elements of $B$ (at most one interval may contain $n$ for any $n \in \mathbb{Z},$ so we do not need to worry about overlap). WLOG, suppose $n=0$ and let $B' = B \cap [0,1].$
For every $b\in B' \subseteq B,$ take the corresponding open interval centered at $b,$ and consider the collection $C$ of all such intervals. By definition, every element of $C$ contains one element of $B';$ denote this property $(*)$. By an analogue of the Bolzano-Weierstrass argument, we have a limit point $b \in B',$ which comes with a sequence $b_1, b_2, \dots \to b.$
Any open interval containing $b$ must contain some $b_i,$ contradicting $(*)$. Thus, $B$ is countable.
The argument for part b can probably be simplified.