Spaces in which all closed sets are regular closed I was reading about the regular closed sets. The definition is 

Let $X$ be a topological space and $A\subseteq X$. We say that $A$ is a regular closed if $A=\text{cl}(\text{int}(A))$

Then, one question came to my mind: is there a topological space $X$ such that $X$ isn't a discrete space and for that every closed subset of $X$ is a regular closed set? 
Obviusly, if $X$ is discrete then every closed set is a regular closed, but, if $X$ isn't discrete, what happens? That example exists? 
Thanks in advance. 
 A: You can take any set $X$ with trivial topology. Then every closed subset in $X$ is trivially regular. 
But if $X$ is $T_1$ and every closed subset is regular then $X$ is discrete. 
A: Given a partition $P$ on a set $X$, we can define a topology whose open sets are unions of elements of $P$.  In this topology, open sets and closed sets are the same, so all closed sets are regular closed.  (If $P$ is the finest partition this is the discrete topology; if $P$ is the coarsest topology it is the indiscrete topology.  Such topologies can also be characterized as the topologies in which closed sets and open sets coincide, or topologies whose $T_0$ quotient is discrete.)
I claim, though, that these are the only examples.  Indeed, suppose $X$ is a topological space in which all closed sets are regular closed.  Suppose $x,y\in X$ are such that $y\in\overline{\{x\}}$.  Since $\overline{\{y\}}$ is regular closed, it is the closure of its interior $U$ which is in particular nonempty, and we must have $y\in U$ since $\{y\}$ is dense in $\overline{\{y\}}$.  Since $y\in\overline{\{x\}}$, we have $U\subseteq \overline{\{x\}}$ as well and so $x\in U$.  Thus $x\in \overline{U}=\overline{\{y\}}$ and so $\overline{\{x\}}=\overline{\{y\}}$.  We see then that $U$ is the interior of $\overline{\{x\}}$ and every element of $\overline{\{x\}}$ is in $U$ (since $y\in U$ and $y$ was originally an arbitrary element of $\overline{\{x\}}$).  Thus $U=\overline{\{x\}}$, so $\overline{\{x\}}$ is open.
So, we have shown that the closure of each singleton in $X$ is a clopen set and is equal to the closure of any of its elements.  It follows easily that the collection of closures of singletons is a partition of $X$, and a subset of $X$ is open iff it is a union of elements of this partition.
