Let $A \subset \mathbb{R}$, $A^\circ$ be the interior of $A$ and $\bar{A}$ be the closure of $A$. Define $\delta A =\bar{A} \setminus A^\circ $. Let $a \in \delta A$.
So far: Let $a \in \delta A$. Then $a \in \bar{A} \setminus A^\circ $. So $a \in \bar{A}$ and $a \notin A^\circ$. Then $a$ is not in the set of interior points of $A$ and $a$ is in the accumulation points of $A$. So by definition, $\exists \epsilon > 0$ such that $(a-\epsilon,a+\epsilon) \subset A$. And also contain infinitely many points $A$. Thus every epsilon neighborhood of $a$ contains a point of $A$. Am I done here? Or is there something that I am missing?
To show that every epsilon neighborhood of $a$ contains a point of $\mathbb{R} \setminus A$. Let $a \in \delta A$. Then $a \in \bar{A} \setminus A^\circ $. So $a \in \bar{A}$ and $a \notin A^\circ$. Then $a$ is not in the set of interior points of $A$ and $a$ is in the accumulation points of $A$. Therefore $\forall \delta > 0$, $(a-\delta,a+\delta)$ is not contained in $A$. Therefore there exists $x \in \mathbb{R}$. Then $x \in \mathbb{R} \setminus A$. Then every epsilon neighborhood of $a$ is contained in $\mathbb{R} \setminus A$