The chromatic number $\chi(X)$ of a topological space $X$ is related to the separation dimension $t(X)$ introduced and studied by [Steinke][1]. The separation dimension $t(X)$ is defined inductively: $\bullet$ $t(\emptyset)=-1$ $\bullet$ $t(X)=0$ for any space $X$ of cardinality $|X|=1$; $\bullet$ if $|X|\ge 2$, then $t(X)\le n$ for some $n\in\mathbb N$ if for each subspace $M\subset X$ with $|M|\ge 2$ there exists a set $A\subset M$ such that $t(A)<n$ and $X\setminus A$ is disconnected. It is easy to see that $t(X)=0$ if and only if the space $X$ is totally disconnected. In Proposition 3.1 of [his paper][1] Steinke proved the following >**Sum Theorem:** For any subspaces $A,B$ of a topological space the union $A\cup B$ has separation dimension $t(A\cup B)\le t(A)+t(B)+1$. This theorem implies that $t(X)+1\le\chi(X)$ for any topological space $X$. On the other hand, by the classical Decomposition Theorem of Urysohn (this is Theorem 7.3.9 in Engelking's book "General Topology"), for a metrizable space $X$ of finite dimension $Ind(X)$ the number $Ind(X)+1$ is equal to the smallest cardinality of a partition of $X$ into subsets of large inductive dimension zero. Since spaces of large inductive dimension zero are totally disconnected, this decomposition theorem implies that $\chi(X)\le Ind(X)+1$ for any metrizable space $X$. Therefore, for any metrizable space $X$ of finite large inductive dimension, we obtain the inequalities: >$$t(X)+1\le \chi(X)\le Ind(X)+1.$$ In Corollary on page 279 of [his paper][1], Stainke proves that for each locally compact paracompact space $X$ we have the inequalities $$dim(X)\le t(X)\le ind(X)\le Ind(X).$$ Since $dim(X)=ind(X)=Ind(X)$ for any separable metrizable space $X$, we finally conclude that >$$dim(X)=t(X)=ind(X)=Ind(X)\quad\mbox{and}\quad\chi(X)=\dim(X)+1$$ for any locally compact separable metrizable space $X$. In particular, >$$\chi(\mathbb R^n)=n+1,$$ which answers the question of N. de Rancourt. [1]: https://link.springer.com/content/pdf/10.1007%2FBF01192781.pdf