The answer is no.

A space is called resolvable if it contains two disjoint dense subspaces. Clearly $X$ is resolvable if and only if $\chi(X)=2$. Lets prove by induction on $n \geq 2$ that if $\chi(X) \leq n$ then $X$ is resolvable (and hence $\chi(X)=2$).

The base case $n=2$ is clear so suppose there is a coloring $f:X \to n+1$. Let $V$ be the union of all open $U$ such that the range of $f\upharpoonright U$ is contained in $n$. Then $V$ is open and $f \upharpoonright V$ is a coloring of $V$ in $n$ colors so that $\chi(V) \leq n$ and by induction hypothesis $V$ is resolvable. Now let $W=X \setminus \mathrm{Cl}(V)$. Since $W$ is open, by definition we have that $f^{-1}(\{n\}) \cap W$ is dense in $W$. But since $f$ is a coloring we also have that $f^{-1}(n) \cap W$ is dense in $W$. This shows that $W$ is also resolvable. Since $V \cup W$ is dense in $X$ we get that $X$ is resolvable.

Note that we didn´t use that $X$ is Hausdorff, connected or anything else (I guess just that $X$ is crowded so that $\chi(X)$ makes sense).

Also note that we can not generalize the above result to infinite cardinals since there are examples of (Tychonoff) countable irresolvable spaces (see for example Alas O., Sanchis M., Tkaˇchenko M.G., Tkachuk V.V., Wilson R.G., Irresolvable and submaximal spaces: Homogeneity versus σdiscreteness and new ZFC examples, Topology Appl. 107 (2000), 259–273.), and such a space must have chromatic number $\aleph_0$.

The answer is no.

A space is called resolvable if it contains two disjoint dense subspaces. Clearly $X$ is resolvable if and only if $\chi(X)=2$. Lets prove by induction on $n \geq 2$ that if $\chi(X) \leq n$ then $X$ is resolvable (and hence $\chi(X)=2$).

The base case $n=2$ is clear so suppose there is a coloring $f:X \to n+1$. Let $V$ be the union of all open $U$ such that the range of $f\upharpoonright U$ is contained in $n$. Then $V$ is open and $f \upharpoonright V$ is a coloring of $V$ in $n$ colors so that $\chi(V) \leq n$ and by induction hypothesis $V$ is resolvable. Now let $W=X \setminus \mathrm{Cl}(V)$. Since $W$ is open, by definition we have that $f^{-1}(\{n\}) \cap W$ is dense in $W$. But since $f$ is a coloring we also have that $f^{-1}(n) \cap W$ is dense in $W$. This shows that $W$ is also resolvable. Since $V \cup W$ is dense in $X$ we get that $X$ is resolvable.

Note that we didn't use that $X$ is Hausdorff, connected or anything else (I guess just that $X$ is crowded so that $\chi(X)$ makes sense).

Also note that we can not generalize the above result to infinite cardinals since there are examples of (Tychonoff) countable irresolvable spaces (see for example Alas O., Sanchis M., Tkačenko M.G., Tkachuk V.V., Wilson R.G., Irresolvable and submaximal spaces: Homogeneity versus σdiscreteness and new ZFC examples, Topology Appl. 107 (2000), 259–273, DOI: 10.1016/S0166-8641(99)00111-X.), and such a space must have chromatic number $\aleph_0$.