A subset $A$ of a topological space $(X,T)$ is said to be semi-open if there exist an open set $B \in ( X,T)$ such that $B \subseteq A \subseteq \overline B$.
Now my question is that
Give an example of a closed set that is not necessarily a semi-open set.
My attempt : I think about closed interval $A=[0,1]$ and $A= \mathbb{R}$ all are satisfied the semi-open properties.
I think such kind example doesn't exist.
On
Let $A$ be a closed set which is not a semi-open set. Then $\overline{A^\circ}\neq A$ (since otherwise it would suffice to pick $B=A^\circ$). Conversely, if $\overline{A^\circ}\neq A$, then for any open subset $B\subset A$, we have $B\subset A^\circ$ (since $A^\circ$ is the maximal open subset of $A$), and therefore $\overline{B}\subset\overline{A^\circ}\subsetneq A$. Hence we have:
Proposition - Let $A$ be a closed set. Then $A$ is a semi-open subset if and only if $\overline{A^\circ}=A$.
Your question can be rephrased as: find closed subsets $A$ such that $\overline{A^\circ}\neq A$; which means $\exists p\in A:p\notin\overline{A^\circ}$; which means $\exists p\in A$ such that $p$ and $A^\circ$ are separated (by open subsets). Suggested by the terminology nowhere dense subset, such a set might be called not dense somewhere. Obviously all non-empty nowhere dense subsets are not dense somewhere (i.e. the case of empty interior).
In the case of $\mathbb{R}$, an isolated point in the set would suffice.
Like quangtu123 said, every closed subset with empty interior does the trick. If you want something in $\Bbb{R}$ which is not just a singleton or something similar, I would suggest looking at the Canot set.