Bourbaki General Topology I: Exercise 20, sec. 8 ch.1 I have added a picture with the complete exercise. I'm interested only in c), I think I have proved a) and b).
In c) we are given a topological space $X_0$ which is semi-regular (i.e. there exists a base for the topology whose open set satisfy the relation $\mathring{\overline{U}}=U$) and another topological space $X$ (the underlying set is always the same). Then, we have to prove that $X^*=X_0$ if and only if there exists a family $\mathfrak M$ of dense subsets of $X$ [According to the exercise, here should be $X_0$ instead of $X$. AR] such that every finite intersection of sets of $\mathfrak M$ belongs to $\mathfrak M$ and such that the topology on $X$ is generated by the union of $\mathfrak M$ and the open sets of $X_0$ (here $X^*$ is the topology whose base are the regular open sets in $X$). Hence, we have to manage to construct examples of topologies finer than regular Hausdorff spaces which are not regular.
To prove the result, the exercise gives us a hint: We should consider the dense open subsets of $X$ and notice that every open set in $X$ can be written as the intersection of a dense open set in $X$ with an open set in $X_0$.
I'm not sure how to start with. I know that, if $D$ is a dense subset, then $\overline U = \overline{U\cap D}$, but I don't think this helps at all.
Any hint will be grateful. Thanks.
EDIT:
The axiom $\mathrm{O_{III}}$ is the condition of regularity: For each closed set $F$ and each point $x\in X\setminus F$, there are disjoint open sets containing $x$ and $F$, respectively.
CONTEXT:
My goal is th goal of the exercise: I want to give finer topologies than regular Hausdorff that aren't regular Hausdorff. I think it is interesting because finer topologies than $T_0$, $T_1$, $T_2$, $T_{21/2}$ or completely Hausdorff are $T_0$, $T_1\dots$, resp. But it doesn't happen for $T_3$, and I would like to know why. My guess is that making finer the topology may appear new closed sets that doesn't verify the $\mathrm{O_{III}}$ axiom; namely, we have made the topology finer, but not enough, so we have create new closed sets but not enough open sets to separatd them from points.
Here is the complete exercise
 A: Your guess is right.
I didn’t find in [B] a definition of a topology $\tau$ generated by a family $\mathfrak N$ of subsets of a set $X$,  so I assume that $\mathfrak N$ is a subbase for $\tau$.

Hence, we have to manage to construct examples of topologies finer than regular Hausdorff spaces which are not regular.
  cone topolgies.

Yes, a simple example is when $X_0$ is a unit segment $[0,1]$ endowed with the natural topology and $\mathfrak M=\{[0,1]\setminus\{1/n:n\in\Bbb N\}\}$.
I was acquainted with this exersise from Bourbaki’s book from your answer, but I applied this construction almost twenty years ago and used it to build Haudorff non-regular paratopological groups, see this my answer and Examples 3 and 2 from [Rav]. This construction turned out to be so basic tool to build counterexamples that later I wrote a paper [Rav2] devoted to its applications.   

$X^*=X_0$ if and only if there exists a family $\mathfrak M$ of dense subsets of $X_0$ such that every finite intersection of sets of $\mathfrak M$ belongs to $\mathfrak M$ and such that the topology on $X$ is generated by the union of $\mathfrak M$ and the open sets of $X_0$.

($\Rightarrow$) Put $\mathfrak M=\{Y: Y$ is open in $X$ and $X\setminus Y$ is nowhere dense in $X_0\}$. Clearly, each set $Y\in\mathfrak M$ is dense in $X_0$. It is easy to check that every finite intersection of sets of $\mathfrak M$ belongs to $\mathfrak M$. Now let $Z$ be any open set of $X$. Let $\overline{Z}$ be the closure of $Z$ in $X$. Since $X^*=X_0$, the interior $Z_0$ of the set $\overline{Z}$ in $X$ is open in $X_0$ and by b) the set $\overline{Z}$ is closed in $X_0$. Let $Y=X\setminus (\overline{Z}\setminus Z)$. It is easy to check that $Y\in\mathfrak M$ and $Z=Y\cap Z_0$.
($\Leftarrow$)  Let $\tau$ be the topology of the space $X_0$ and $\sigma$ be the topology on the set $X$ with the subbase (in fact, a base) $\mathfrak M$. The topology of the space $X$ is a supremum $\tau\vee\sigma$ of topologies $\tau$ and $\sigma$. It is easy to see that the topologies $\tau$ and $\sigma$ are cowide and the topology $\sigma$ is wide, see definitions on [Rav2, p.10]. Since the topology $\tau$ is semiregular, $\tau_r=\tau$ (see [Rav2, p.11]) and by [Rav2, Lemma 7], $(\tau\vee\sigma)_r=\tau_r=\tau$, that is $X^*=X_0$.  
References 
[B] Nicolas Bourbaki, Elements of mathematics. General topology 1,  Springer, 1966?.
[Rav] Alex Ravsky, *Pseudocompact paratopological groups , version 5.
[Rav2] Alex Ravsky, Cone topologies of paratopological groups.
