Closure of interior of closed set

If $D$ is a closed set, what is the relation in general between the set $D$ and the closure of $\operatorname{Int}D$?

We know that $\operatorname{Int}D\subseteq D$, so $\overline{\operatorname{Int}D}\subseteq \overline{D}$, but since $D$ is closed, we have $\overline{D}=D$, so that $\overline{\operatorname{Int}D}\subseteq D$.

Now, is it true as well that $D\subseteq \overline{\operatorname{Int}D}$? I can't seem to prove it, or give an example of $D$ such that this doesn't hold.

Hint: For a counterexample, try to think of a non-empty closed set with empty interior.

In general, the other inclusion doesn't hold. For example, if $D = \{0\}$ is the set containing the single point 0, then its interior is empty. There are a lot of closed sets with this property, like finite sets and Cantor sets.

Since $\operatorname{Int}D\subseteq D$, obviously $\overline{\operatorname{Int}D}\subseteq \overline{D}=D$ holds.