Let $Q$ be a fixed subset of $X$.
Define $\tau = \{Y \subseteq X | Q \subseteq Y\ \} \cup \emptyset$
I already proved that this makes $(X,\tau)$ a topological space verifying the three axioms of the definition of topology.
Now I am supposed to say under what conditions this topological space is regular. And also completely regular.
So for the regular case, I considered $x \in X$ and $F \subseteq_{cl}X$ and am supposed to say what would make possible the existence of $A,B \in \tau$ disjoint such that $x \in A$ and $F \subseteq B$.
$X \setminus F \in \tau$, then $Q \subseteq X \setminus F$, then $F \subseteq X \setminus Q$. But I am not sure what to say next. How can I proceed? I also tried using this theorem: $X$ is regular $\iff \forall U \in \tau, x \in U$ there is $V \in \tau$ such that $x \in V, \overline{V} \subseteq U$. But didn't get any useful conclusion.
Any help would be appreciated.
If $Q= \emptyset$ the $\tau$ is the discrete topology i.e. each set is open hence $X$ is regular. Otherwise each pair of open set $A$ and $B$ clearly have not trivial interception ($Q\subset A\cap B$) so that $X$ cannot be regular