What happens if $U\in \textit{P}(X)$ is always either closed or open Let $X$ be a topological space s.t. $\forall U\in \textit{P}(X)$, $U$ is either open or closed (or both only in the cases $U=X, U=\varnothing$).
What can we say about $X$?
I found this example of such a space: Let $X$ a set. Fix $u\in X$, and then say that $U\in \textit{P}(X)$ is closed $\iff$ $u\in U$ or $U=\varnothing$.
Note that in the above example I could substitute "open" with "closed" and it would still work.
My conjecture (which is poorly founded) is that all such topological spaces are of the above forms.
 A: This thread is about so-called "door spaces" which is the name for a topological space where every subset is open or closed (like a door). It gives a link to a paper that proves there are three types of connected door spaces: included point topologies (there is a point so that $O$ is open iff $p \in O$ or $O$ is empty), excluded point topologies (the same with closed instead of open, as you name) and one based of a free (non-principal) ultrafilter on $X$. These objects only exist under a form of the axiom of choice (they're not "constructive").
A: Mark Kamsma's comment answers your question. I figured I'd expand on it, since you seem to be ignoring it.
Let $F$ be an ultrafilter on a set $X$. Recall the definition of an ultrafilter on a set. An ultrafilter on $X$ is a nonempty collection $F$ of subsets of $X$ such that

  
*
  
*$\varnothing\not \in F$,
  
*If $A\in F$ and $A\subseteq B$, then $B\in F$.
  
*If $A,B\in F$, then $A\cap B\in F$.
  
*For all $A\subseteq X$, either $A\in F$ or $A^C\in F$.
  

Then the collection $\tau = F\cup \{\varnothing\}$ is a topology for $X$ satisfying the condition that for every nonempty proper subset $\varnothing \subsetneq U\subsetneq X$, exactly one of $U$ and $U^C$ is in $\tau$.
Being a topology follows from properties 2 and 3 of $F$ being an ultrafilter, and satisfying the property you want follows from properties 1 and 4. 
Thus a nonprincipal ultrafilter, produces a topology which is a counterexample to your conjecture.
