How to show that $\Bbb R$ with the topology generated by $\{(a,b):a,b\in\Bbb R\}\cup\{(a,b)\cap\Bbb Q:a,b\in\Bbb R\}$ is connected? I'm sorry that I'm not good at English. My question is

Let $B = \{(a,b) : a,b \in \mathbb{R} \} \cup \{ (a,b) \cap \mathbb{Q} : a,b \in \mathbb{R} \}$ is basis of $\tau$. Then $(\mathbb{R}, \tau)$ is connected?

I think $(\mathbb{R}, \tau)$ is connected.
Because form of open set is finite intersection of $(a,b)$ or $(a,b) \cap \mathbb{Q}$, arbitrary union of $(a,b)$ or $(a,b) \cap \mathbb{Q}$.
Form of closed set is finite union of $[a,b]$ or $[a,b]-\mathbb{Q}$ or arbitrary intersection of $[a,b]$ or $[a,b]-\mathbb{Q}$.
So closed and open set is either $\emptyset$ or $\mathbb{R}$. But i don't know how to show that this. Help me. Thank you for reading what i wrote .
 A: Note the following property fo $\tau$:

If $O$ is open and $x\in O$, then there exist $a,b\in\Bbb R$ with $a<x<b$ and $(a,b)\cap\Bbb Q\subseteq O$.

Assume $\Bbb R=U\cup V$ with $U,V$ non-empty open. We have to show that $U\cap V\ne\emptyset$.
Pick $u\in U$, $v\in V$. Wlog. $u<v$.
Let $$w=\inf\{\,x\in \Bbb R\mid x>u\land \exists y>x\colon (x,y)\cap\Bbb Q\subseteq V\,\}$$
(so that $u\le a<v$). 
If $w=u$, we are done.
So we can assume $w>u$.
If $w\in V$, then by the property in the first paragraph, $(a,b)\cap \Bbb Q\subseteq V$ for some $a<w<b$. If $a>u$, we arrive at a contradiction with the definition of $w$ as an infimum. And if $a\le u$, then also $(\frac{u+w}2,b)\cap \Bbb Q\subseteq V$ and we arrive at the same contradiction.
We conclude that $w\notin V$, hence $w\in U$.
Therefore, there exist $a<w<b$ with $(a,b)\cap \Bbb Q\subseteq U$. By definition of infimum, there exist $x,y$ with $u<x<b$ and $(x,y)\cap \Bbb Q\subseteq V$. Then the non-empty set $(x,\min\{y,b\})\cap \Bbb Q$ is $\subseteq U\cap V$. $\square$
