The family $\mathfrak{B}= \tau \cup \{\{x \} \mid x \in \Bbb P\}$ is a basis for some topology $\pi$ of $\Bbb R$. Show that $(\Bbb R, \pi)$ is regular 
Let $\tau$ be the standard topology on $\Bbb R$ and let $\Bbb P = \Bbb R \setminus \Bbb Q$. The family $\mathfrak{B}= \tau \cup \{\{x \} \mid x \in \Bbb P\}$ is a basis for some topology $\pi$ of $\Bbb R$. Show that $(\Bbb R, \pi)$ is regular.

To show regularity I'm used to using the characterization that if $x \in X$ and $U$ is a open set containing $x$, then if for $V$ open set containing $x$ we have that $\overline{V} \subset U$ the space is regular.
So trying to use this if $x \in \Bbb R$ and $U$ is a open set containing $x$ I know that there is a basis set $B \in \mathfrak B$ such that $$x \in B \subset U$$
There are only two possibilities for $B$ either it's in $\tau$ or it's an irrational singleton. If $B \in \tau$ I think I can use the fact that $(\Bbb R, \tau)$ is regular, but I don't know wheter it will work as I'm taking the closures w.r.t to the topology $\pi$?
Also if $B$ is a singleton I think I run into some issues as $x$ can be any real number not irrational neccessarily?
 A: To begin understanding what's going on here, notice that your open sets are basically the usual opens from $\mathbb{R}$ together with singletons consisting of irrational numbers. This has an interesting consequence: singletons of irrational numbers are now both open and closed, because every singleton is already closed with respect to the usual topology $\tau$. The topology $\pi$ also adds a bunch of new closed sets to the topology, namely every set of the form $\mathbb{R}\setminus\{x\}$, where $x \in \mathbb{P}$. But as we already discussed, every $\{x\}$ with $x \in \mathbb{P}$ is both open and closed, so every corresponding complement $\mathbb{R}\setminus\{x\}$ is also open and closed.
Now, to answer your question note that a different characterization of a regular is space is that given a closed set $C$ and a point $x$ not contained in $C$, there are open neighbourhoods of $C$ and $x$ which do not intersect. That is, $C \subset U, x \subset V$, where $U, V$ are open and $U \cap V = \emptyset$.
If $C$ is closed and $x \in \mathbb{R}\setminus C$, then either $C$ is a closed with respect to $\tau$, in which case regularity follows because $(\mathbb{R}, \tau)$ is already regular, or $C$ is one of the new closed sets introduced by $\pi$. In the second scenario, there are two options. Either $C$ is simply the complement of a irrational singleton (that is, a base element), or $C$ is an arbitrary intersection of such complements. I'll treat them separately to make the idea easier to understand.
In the first case we have $C = \mathbb{R}\setminus\{x\}$ for some irrational $x$. But in this case, the only possible point outside $C$ is $x$, and we can choose the disjoint open neighbourhoods to be $C$ and $\{x\}$, since these are both open (and closed) in $\pi$.
In the second case, $C = \cap_\alpha (\mathbb{R}\setminus\{x_\alpha\}) = \mathbb{R}\setminus\{x_\alpha\}_\alpha $, where $x_\alpha$ is irrational. Again, a point outside $C$ must be one of the irrational numbers in the family $\{x_\alpha\}_\alpha$, say $x_\beta$, but now is not clear whether $C$ is open or not, since it is an arbitrary intersection of open sets. We know, however, that $C \subset \mathbb{R}\setminus\{x_\beta\}$, which is open. So we choose as neighbourhoods the open sets $\mathbb{R}\setminus\{x_\beta\}$ and $\{x_\beta\}$.
A: Case I: if $x\in \mathbb Q$ then an open set $U$ containing $x$ contains a basis element of the form $(a,b):a,b\in \mathbb Q.$ Choose irrational numbers $x<y$ such that $a<x<y<b.$ Then, $[x,y]$ is open in the new topology (it is a union of the open sets $\{x\},\{y\}$ and $(a,b))$ and $[x,y]\subseteq (a,b)\subseteq U.$ But $[x,y]$ is also closed in the new topology (because its complement is open.)
Case II: if $x\in \mathbb P$ then any open set $U$ containing $x$ satisfies $x\in \{x\}=\overline{\{x\}}\subseteq U.$
