Question: Let $X \subset \mathbb{R}$, then the set $$ A := \{ x \in X \mid \exists \delta > 0 \; s.t. (x-\delta,x) \cap X = \emptyset \}$$ is at most countable.
My thought: Suppose $A$ is uncountable, then prove that there exists an open interval $B \subset A$, then we get the contradiction. But I don't know how to find such open interval $B$. Maybe my thought doesn't work. Who could help me with this question?
For each $x \in A$, fix a $\delta_x > 0$, s.t. $(x-\delta_x,x) \cap X = \emptyset$, then fix a rational $r_x \in (x-\delta_x,x) $. Define $f: A \longrightarrow \mathbb{Q}$ by $f(x)=r_x$, then check it is injective: Let $ x \neq y \in A$, WLG, $x < y$, since $(y-\delta_y,y) \cap X= \emptyset$, then $x \notin (y-\delta_y,y)$, then $x \leq y-\delta_y$, so $(x-\delta_x,x) \cap (y-\delta_y,y) = \emptyset$, hence $r_x \neq r_y$. So $f$ is injective.