Connectedness of the topological space on $\mathbb{R}$ with topology generated by usual topology and the set of irrationals 
Let $\mathcal{U}$ be the usual topology on the Euclidean line $\mathbb{R}$. Let $\mathcal{T}$ be the topology generated by $\mathcal{U} \cup \left\{ U \setminus \mathbb{Q}| U \in \mathcal{U} \right\}$. Determine if the topological space $(\mathbb{R}, \mathcal{T})$ is connected.

I was trying to make a variation of the K topology. The philosophy of the K topology is to make a topology finer by attaching a non-open set as a subbasis element. So I figure out the above one expecting to become an interesting and 'easy' example.
However, after some trial, I failed to prove or disprove the connectedness of the space.
I guess it is connected. I began, as usual, letting $U$ be a non-trivial clopen set containing $a$ but not $b$ where $a<b$ where generality is not lossed, and tried argue to show that the lower limit $l$ of $(a,\infty) \setminus U$ is a contradictory element. However, it was not that easy.
Pleas help to improve this problem with an illumination solution!
 A: Suppose $U,V\in \mathcal{T}$ and $U\cup V=\mathbb{R}$. We will show that $U\cap V\neq \emptyset$.
We have  $$U=U_1\cup (U_2\cap I),\qquad V=V_1\cup (V_2\cap I),$$ where $I=\mathbb{R}\backslash\mathbb{Q}$ denotes the irrationals, and   $U_1,U_2,V_1,V_2\in \mathcal{U}$.
As $U\subseteq U_1\cup U_2$ and $V\subseteq V_1 \cup V_2$, we have that $U_1\cup U_2$ and $V_1\cup V_2$ together cover $\mathbb{R}$ and are open in the usual topology, so $$(U_1\cup U_2)\cap(V_1\cup V_2)\neq\emptyset.$$
This non-empty open set in the usual topology must contain an interval, which must contain some $x \in I$.  Thus $x\in I\cap(U_1\cup U_2)\cap(V_1\cup V_2)$:
$$
x\in U_1\cup(U_2\cap I)=U,\qquad x\in V_1\cup(V_2 \cap I)=V.
$$
Hence  $x\in U\cap V$ and $U\cap V\neq \emptyset$.  We conclude $(\mathbb{R},\mathcal{T})$ is connected.
A: $\newcommand{\cl}{\operatorname{cl}}$HINT: Suppose that $U\in\mathcal{T}$ is clopen with respect to $\mathcal{T}$, and $\varnothing\ne U\ne\Bbb R$.

*

*Verify that $U$ can be written as $U=V\cup(W\setminus\Bbb Q)$ for some $V,W\in\mathcal{U}$.

*Show that it $q\in W\cap\Bbb Q$, then $q\in\cl_{\mathcal{T}}W\subseteq\cl_{\mathcal{T}}U=U$.

*Conclude that $U=V\cup W\in\mathcal{U}$.

*Show that there is an $x\in(\cl_{\mathcal{U}}U)\setminus U$.

*Show that $x\in\cl_{\mathcal{T}}U$ to get a contradiction.

You may find it useful to verify that the irrationals are still dense in $\langle\Bbb R,\mathcal{T}\rangle$.
