The chromatic number $\chi(X)$ of a topological space $X$ is related to the separation dimension $t(X)$ introduced and studied by Steinke.
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 $M\setminus A$ is disconnected;
$\bullet$ $t(X)=n$ for some integer $n\ge 0$ if $t(X)\le n$ and $t(X)\not\le n-1$.
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 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, 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, we obtain the following theorem answering the question of N. de Rancourt.
Theorem 1. For every $n\in\mathbb N$ the Euclidean space $\mathbb R^n$ has chromatic number $\chi(\mathbb R^n)=n+1.$
For general separable metrizable spaces, we have the following upper bound, which can be interesting for Set Theorists.
Theorem 2. Each separable metrizable space $X$ has chromatic number $\chi(X)\le\omega_1$.
Proof. Choose a family $(D_\alpha)_{\alpha\in\omega_1}$ of pairwise disjoint dense sets in the real line $\mathbb R$. For every countable ordinal $\alpha$ consider the set $Z_\alpha=\mathbb R\setminus\bigcup_{\alpha\le\beta<\omega_1}D_\beta$ and observe that $(Z_\alpha)_{\alpha\in\omega_1}$ is an increasing transfinite seqeunce of zero-dimensional subspaces of $\mathbb R$ such that $\bigcup_{\alpha\in\omega_1}Z_\alpha=\mathbb R$.
Taking into account that the cardinal $\omega_1$ has uncountable cofinality, we can show that $\{Z_\alpha^\omega\}_{\alpha<\omega_1}$ is a cover of $\mathbb R^\omega$ by $\omega_1$ many zero-dimensional subspaces, which yields the upper bound $\chi(\mathbb R^\omega)\le\omega_1$.
Since each separable metrizable space embeds into $\mathbb R^\omega$, we finally obtain the desired upper bound $\chi(X)\le\chi(\mathbb R^\omega)\le\omega_1$.
This upper bound is attained for the Hilbert cube.
Theorem 3. The Hilbert cube $\mathbb I^\omega=[0,1]^\omega$ has chromatic number $\chi(\mathbb I^\omega)=\omega_1$.
Proof. The upper bound $\chi(\mathbb I^\omega)\le\omega_1$ was proved in Theorem 1 and the lower bound $\chi(\mathbb I^\omega)>\omega$ was proved by Krasinkiewicz (see also this paper).