Let $A$ be a domain and $K=\mathrm{Frac}(A)$.

The Zariski Riemann space $\mathrm{ZR}(K,A)$ is the set of all valuation rings of $K$ containing $A$. It comes with a natural center map \begin{align}c_A:\mathrm{ZR}(K,A)&\rightarrow \mathrm{Spec}\,A \\ (\mathcal{O},\mathfrak{m})&\mapsto\mathfrak{m}\cap A\end{align}

Problem:Let $\mathcal{O}_1, \mathcal{O}_2$ valuation rings in $\mathrm{ZR}(K,A)$ such that $\mathcal{O}_1 \subsetneq \mathcal{O}_2$. Is it true that $c_A(\mathcal{O}_1)\subsetneq c_A(\mathcal{O}_2)$?

As $\mathcal{O}_1 \subseteq \mathcal{O}_2\iff\mathcal{O}_1 \in \overline{\{\mathcal{O}_2\}}$ with the Zariski topology of $\mathrm{ZR}(K,A)$, and as every irreducible closed subset is the closure of a point (because $\mathrm{ZR}(K,A)$ is a Spectral space). This can be used to understand how $c_A$ relates the naive Krull dimension definition on $\mathrm{ZR}(K,A)$ to the Krull dimension of $\mathrm{Spec}(A)$.

Also notice that if $\mathcal{O}_1 \subsetneq \mathcal{O}_2$ then $\mathcal{O}_2$ is the localization of $\mathcal{O}_1$ on the prime $\mathfrak{m}_2\in \mathrm{Spec}\,\mathcal{O}_1$. Hence, it is equivalent to prove that the map $\mathrm{Spec}\,\mathcal{O}_1\rightarrow \mathrm{Spec}(A_{\mathfrak{m}_1\cap A})$ is injective. So we can restate the problem as

Problem':Let $A$ be a local domain dominated by a valuation ring $\mathcal{O}$ of $\mathrm{Frac}(A)$. Does the extension of rings $A\subseteq \mathcal{O}$ satisfy the incomparability property?

**Edit later:** Here is another more geometric counterexample on algebras of finite type. Let $X$ be the blow-up of $\mathbb{A}^2$ at $(0,0)$ and let $E$ be the exceptional divisor. By taking a point in $p\in E$ we can define a rank $2$ valuation in $K=K(X)$ that sends a rational function $f$ to $(n_1,n_2)$ where $n_1$ is the order of vanishing at $E$ and $n_2$ is the order of vanishing of $(f\pi^{-n_1})_{|E}$ at $p$ where $\pi$ is a local equation for $E$ around $p$. If $\mathcal{O}$ is the valuation ring of this valuation then it dominates $A=\mathcal{O}_{\mathbb{A}^2,(0,0)}$ but the two primes ideals of it goes to the maximal ideal in $A$.