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 thatBut in fact $\chi_{\mathrm{cf}}(\Bbb R)>2$, indeed suppose$\aleph_0$ is a lower bound in any reasonable space.
Lemma 1: Let $X$ be a $T_1$ space with at least three points. Then $\chi_{\mathrm{cf}}(X)>2$.
Proof: Suppose for a contradiction that it were $2$ and let$\chi_{\mathrm{cf}}(X)=2$ as witnessed by $c\colon\Bbb R\to 2$ witness it$c\colon X\to 2$. Let $U$ be a nonempty open set, then there must$x\in X$ be a point $u\in U$ withsuch that $c(u)\neq c(u')$$c(x)\neq c(x')$ for everyall $u'\in U$$x'\in X\setminus\{x\}$. But now we have that $U\setminus\{u'\}$$X\setminus\{x\}$ is a monochromatic open set in $X$ with at least two points, a contradiction. By induction we see that
Lemma 2: Let $\chi_{\mathrm{cf}}(\Bbb R)$ cannot$X$ be finitean infinite $T_1$ space. Indeed supposeThen $\chi_{\mathrm{cf}}(\Bbb R)=n$ and let$\chi_{\mathrm{cf}}(X)\geq\aleph_0$.
Proof: Suppose for a contradiction $c\colon\Bbb R\to n$ witness it$\chi_{\mathrm{cf}}(X)=n$ as witnessed by $c\colon X\to n$. FindLet $r\in\Bbb R$ so$x_1\in X$ be such that $c(r)\neq c(r')$$c(x)\neq c(x')$ for everyall $r'\in\Bbb R$$x'\in X\setminus\{x_1\}$. But nowNow $c\upharpoonright\Bbb R\setminus\{r\}$$c\upharpoonright X\setminus\{x_1\}$ witnesses that $\chi_{\mathrm{cf}}(\Bbb R\setminus\{r\})\leq n-1$, but this$\chi_{\mathrm{cf}}(X\setminus\{x_1\})\leq n-1$. Proceed inductively to find $x_1,\ldots,x_{n-2}$ so that $c\upharpoonright X\setminus\{x_1,\ldots,x_{n-2}\}$ witnesses that the latter space contains a homeomorphic copy ofhas conflict-free chromatic number at most $\Bbb R$$2$, giving a contradictionwhich contradicts Lemma 1.
In conclusion $\chi_{\mathrm{cf}}(\Bbb R)=\aleph_0$.