... actually, even more general statement is true:
Theorem. Every regular ring is universally catenary.
(see for example Algebraic Geometry by Qing Liu, Corollary 2.16, Chapter 8)
Though, the proof is neither easy nor elementary. On the other hand, there should exist an easy proof for Dedekind domains, or at least for principal ideal domains. I only managed to reduce the problem to the following property:
Let $A$ be a principal ideal domain and $B$ a finitely generated over-domain. Then the dimension of each fibre over a closed point is equal to the dimension of the generic fibre. So in other words $$ \dim B\otimes_A A_{(0)}=\dim B\otimes_A k(\mathfrak{p}) $$ for every nonzero prime ideal $\mathfrak{p}\lhd A$.
The last statement follows easily from dimension formula, so a posteriori we know it's true. I am wondering if there exist some independent proof.
I am not sure what exactly your questions is. But here I am assuming the fact that a regular ring is catenary.
Let $A$ be a regular ring. $B$ a finitely generated $A$-algebra, i.e., $B = A[x_1,\dots, x_n]/I$. We first show that it suffices to prove the statement for polynomial ring (finitely many variables) over $A$. Let $p \subseteq q$ be prime ideals in $B$. Since there is a 1-1 correspondence between the prime ideals in $A[x_1,\dots, x_n]$ contain $I$ and the prime ideals in $B$, it suffices to prove the statement for the pre-images of $p,q$ in $A[x_1,\dots,x_n]$. By our assumption it is catenary. Therefore, saturated chains of primes ideal between $p, q$ have the same length.
Since $A$ is regular a polynomial ring over $A$ is also regular. Therefore, it is catenary.