By colouring every element of $\Bbb Q$ with a unique colour and every element of $\Bbb R\setminus\Bbb Q$ with the same colour (different from all the colours used so far), we see that $\chi_\mathrm{cf}(\Bbb R)\leq\aleph_0$.

It's easy to see that $\chi_{\mathrm{cf}}(\Bbb R)>2$, indeed suppose for a contradiction that it were $2$ and let $c\colon\Bbb R\to 2$ witness it. Let $U$ be a nonempty open set, then there must be a point $u\in U$ with $c(u)\neq c(u')$ for every $u'\in U$. But now $U\setminus\{u'\}$ is a monochromatic open set, a contradiction. By induction we see that $\chi_{\mathrm{cf}}(\Bbb R)$ cannot be finite. Indeed suppose $\chi_{\mathrm{cf}}(\Bbb R)=n$ and let $c\colon\Bbb R\to n$ witness it. Find $r\in\Bbb R$ so that $c(r)\neq c(r')$ for every $r'\in\Bbb R$. But now $c\upharpoonright\Bbb R\setminus\{r\}$ witnesses that $\chi_{\mathrm{cf}}(\Bbb R\setminus\{r\})\leq n-1$, but this space contains a homeomorphic copy of $\Bbb R$, giving a contradiction.

In conclusion $\chi_{\mathrm{cf}}(\Bbb R)=\aleph_0$.