How can I show that this space is regular Consider the collection of all sets of the form $U \cup (V\cap \Bbb{Q})$ where $U,V$ are open in standard topology on $\Bbb{R}$. I have shown that  such a collection is a topology which I call $\mathcal{T}$.
Now I can show that $\mathcal{T}$ is Hausdorff. I also know that $\Bbb{Q}$ is open in this topology and furthermore that there is no open set that completely contains in $\Bbb{Q}^C$. But how can I show that this space is not regular?
It seems like right now I am an eagle circulating over fish in the ocean but somehow am unable to plunge in to pick the fish. 
Please don't give it all away.
 A: Hint: The closure of every nonempty open set in this space must contain irrationals (given $A \subseteq \mathbb{R}$ and any irrational number $x$ one can show that $x \in \overline{A}$ iff $x \in \mathrm{cl}_{\text{met}} (A)$, where the latter is the closure in the usual metric topology), but $0$ has open neighbourhoods which contain no irrational numbers.
A: You are correct in having shown that $\Bbb{Q}^c$ is closed. Now consider the point $0$. Then $\{0\} \cap \Bbb{Q}^c=\varnothing$ and furthermore we claim that these can't be separated as follows.
Suppose they can.
Any open set containing $\Bbb{Q}^c$ is of the form $U \cup (V \cap \Bbb{Q})$ where $U,V$ are open in the Euclidean topology. Now if $x \in \Bbb{Q}^c$ then $x$ cannot be in $V \cap \Bbb{Q}$ and so $\Bbb{Q}^c \subseteq U$.
On the other hand any open set about $0$ can't contain irrationals and so is of the form $W \cap \Bbb{Q}$ where $W$ is open in the Euclidean topology. But now
$$W \cap \Bbb{Q} \cap U$$
is non-empty showing our claim.
Exercise: Prove that $W \cap \Bbb{Q} \cap U$ is non-empty.
