A set $A\subseteq\mathbb{R}$ is nowhere dense if for any interval $I,$ there exists a subinterval $J\subseteq I$ such that $J \subseteq A^c.$
$x\in A$ is an interior point if there exists an open interval $I$ of $x$ such that $I\subseteq A.$
We denote $Int(A)$ as the set of all interior points of $A.$
Statement: If $A$ is nowhere dense, then $$Int(\overline{A})=\emptyset.$$
My attempt: Suppose that $$Int(\overline{A})\neq\emptyset.$$ Let $x\in Int(\overline{A}).$ By definition of interior, there exists an open interval $I$ of $x$ such that $I\subseteq \overline{A}.$
By definition of nowhere dense of $A,$ there exists a subinterval $J\subseteq I$ such that $J\subseteq A^c.$ Then we have $$J\subseteq \overline{A}\cap A^c.$$ I do not know how to proceed from here. I think there is a contradiction, but I fail to see it. Any hint will be appreciated.
EDITED: $x\in \overline{A}$ if for any open interval $I$ containing $x,$ we have $I \cap A \neq \emptyset.$
(I). If $J$ is an open set then for any $A$ we have $J\cap \overline A=\phi \iff J\subset A^c.$
(II). If $Int (\overline A)\ne \phi$ there exists a non-empty open interval $I\subset \overline A.$ Then no non-empty open interval $J\subset I$ can be a subset of $A^c.$ Otherwise $J\cap \overline A=\phi$, implying $$\phi=J\cap \overline A =(J\cap I)\cap \overline A=J\cap (I\cap \overline A)=J\cap I=J\ne \phi$$ which is absurd.
So if $Int (\overline A)\ne A$ then $A$ is not nowhere-dense.
(III). If $Int(\overline A)=\phi$ and $I$ is any non-empty open set then $I\setminus \overline A$ is a non-empty open set so there exists a non-empty open interval $J\subset (I\setminus \overline A).$ Since $J$ is open and disjoint from $\overline A$ we have $J\subset A^c.$
So if $Int(\overline A)=\phi$ then $A$ is nowhere-dense.