Prove that if $\mathcal{O}$ is open, then the interior of $\bar{\mathcal{O}} - \mathcal{O}$ is empty I am trying to show the following:

Let $\mathcal{O}$ be an open subset of a metric space $X$. Show that
$\text{int} \, (\bar{\mathcal{O}} - \mathcal{O}) = \emptyset$.

Here, for $E \subset X$, $\text{int} \, E$ denotes the interior of a set and $\bar{E}$ denotes its closure in $X$.
My attempt:
For the sake of contradiction, suppose $\bar{\mathcal{O}} - \mathcal{O}$ has non-empty interior, so there is $x \in \text{int} \, (\bar{\mathcal{O}} - \mathcal{O})$. This means there is $r > 0$ such that $B(x, r) \subset \bar{\mathcal{O}} - \mathcal{O}$. However, since $x$ is a point of closure of $\mathcal{O}$, for every $t > 0$, $B(x, t) \cap \mathcal{O} \neq \emptyset$. This contradicts with $B(x, r) \subset \bar{\mathcal{O}} - \mathcal{O}$. Hence $\text{int} \, (\bar{\mathcal{O}} - \mathcal{O})$ is empty. $\blacksquare$
The problem I have with my attempt is that the assumption of $\mathcal{O}$ being open is not used. The argument seems to work for any subset of $X$, or in other words, it suggests that if $E$ is any subset of $X$, then $\text{int} \, (\bar{E} - E) = \emptyset$. I do not think this is true, but I have not came up with a counter-example to debunk it.
Please help by:

*

*spotting any mistakes in my attempt;

*answering the question " for $E \subset X$, does $\text{int} \, (\bar{E} - E) = \emptyset$ holds for all $E \subset X$ "?

Any help is appreciated.

Here are the definitions I used: For $E \subset X$, $x \in X$ and $r > 0$,

*

*$B(x, r)$: open ball centered at $x$ with radius $r$.

*$\text{int} \, E$: interior of $E$, the points in $E$ which there is an open ball centered at $x$ covered by $E$, i.e.
$$
  \text{int} \, E = \{ x \in E : \exists r > 0 \, , B(x, r) \subset E \} \ .
$$

*$\bar{E}$: closure of $E$, the points in $X$ which every open set containing $x$ also contains a point in $E$, i.e.
$$
  \bar{E} = \{ x \in X: \forall \text{ open set } O \subset X \text{ such that }x \in O \, , O \cap E \neq \emptyset\} \ .
$$
 A: The proof is fine to the point that it can be easily generalised to an arbitrary topologial space as shown below.

Definition
Let be $X$ a topological space. A set $U$ is a neigborhood of any $x\in X$ if there exist an open set $V$ such that
$$
x\in V\subseteq U
$$
Definition
Let be $X$ a topological space. The interior $\operatorname{int} A$ of any subset $A$ of $X$ is the union of all open set contained in $A$.
Definition
Let be $X$ a topological space. A point $x$ is a cluster point of $A$ if any neigborhood of $x$ contains any point of $A$.
Definition
Let be $X$ a topological space. The clousure $\operatorname{cl} A$ of a subset $A$ of $X$ is the subset of $X$ which contains all points of $A$ and of cluster points of $A$.

So with respect this definitions replacing $B(x,r)$ with a generic neighborhood of $x$ with the same arguments you gave it is possible to show that if $A$ is any subset of $X$ then
$$
\operatorname{int}\big(\operatorname{cl}A\setminus A\big)=\emptyset
$$
A: Your proof is indeed correct for general $O.$ Nice! I'll provide you with a different version, because, why not.
Let $V$ be the interior of $\bar{O} \setminus O$. Then $V$ is open and a subset of $\bar{O}$ and since the interior of $\bar{O}$ is the same as that of $O,$ namely, $\mathring{O},$ and the interior of a set is the largest open subset, we get $V\subseteq \mathring{O} \subseteq O.$
On the one hand, we have $V \subseteq \bar{O} \setminus O.$ On the other, we have $V\subseteq O.$ Therefore $V = \emptyset.$
A: Let $x$ be an interior point of $\mathcal{\overline{O} } -\mathcal{O } $ then there should be a neighbourhood of $x$ in $\mathcal{\overline{O} } -\mathcal{O }$. But since $x$ is a limit point of $\mathcal{O } $, such neighbourhood must contains points of $\mathcal{O } $, which is impossible.
