If τ is the finite-closed topology, then the semi-open sets are precisely the open sets

Question

If τ is the finite-closed topology, then the semi-open sets are precisely the open sets.

We can assume the set is infinite, otherwise the proof is trivial. I can easily show that open sets are semi-open. Any open set A are element of X without a finite number of points. If we remove another point, it is still an open set, O, and is a subset of A. The closure of O is the entire set because that is the smallest closed set that contains O since every closed set is finite. So every open set fits the criterion for being semi open. Now I need to show that if there are any other semi open sets in the topology, is is just part of the open sets. I am not sure how to show this.

Answer 1

Let $X$ be an infinite space with the cofinite topology. Let $A\subseteq X$ be semi-open, meaning that there is an open $U$ in $X$ such that $U\subseteq A\subseteq\operatorname{cl}U$. If $U=\varnothing$, then $\operatorname{cl}U=\varnothing$, so $A=\varnothing$, which is open. If $U$ is non-empty, then $U=X\setminus F$ for some finite $F\subseteq X$, and $\operatorname{cl}U=X$, so $X\setminus F\subseteq A\subseteq X$. But then $X\setminus A\subseteq F$, so $X\setminus A$ is finite, and $A$ is open.