First, let me explicitly assume $R$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$.
Proposition. The Krull dimension of $R$ is at most $1$.
Proof. Let $\mathfrak{m} = (t)$, and let $\mathfrak{p}$ be prime. We know $\mathfrak{p} \subseteq \mathfrak{m}$, so it is enough to show that either $\mathfrak{p} = \mathfrak{m}$ or $\mathfrak{p} = (0)$. Suppose $\mathfrak{p} \ne \mathfrak{m}$: then $t \notin \mathfrak{p}$. Let $a_0 \in \mathfrak{p}$. For each $a_n$, because $\mathfrak{p}$ is prime, there exists $a_{n+1}$ in $\mathfrak{p}$ such that $a_n = a_{n+1} t$. By the axiom of dependent choice, this yields an infinite ascending sequence of principal ideals:
$$(a_0) \subseteq (a_1) \subseteq (a_2) \subseteq \cdots$$
Since $R$ is noetherian, for $n \gg 0$, we must have $(a_n) = (a_{n+1}) = (a_{n+2}) = \cdots$. Suppose, for a contradiction, that $a_0 \ne 0$. Then, $a_n \ne 0$ and $a_{n+1} \ne 0$, and there is $u \ne 0$ such that $a_{n+1} = a_n u$. But then $a_n = a_{n+1} t = a_n u t$, so cancelling $a_n$ (which we can do because $R$ is an integral domain), we get $1 = u t$, i.e. $t$ is a unit. But then $\mathfrak{m} = R$ – a contradiction. So $a_n = 0$. $\qquad \blacksquare$
Here's an elementary proof which shows why we can reduce to the case where $R$ is an integral domain.
Proposition. Any non-trivial ring $A$ has a minimal prime.
Proof. By Krull's theorem, $A$ has a maximal ideal, which is prime. Let $\Sigma$ be the set of all prime ideals of $A$, partially ordered by inclusion. The intersection of a decereasing chain of prime ideals is a prime ideal, so by Zorn's lemma, $\Sigma$ has a minimal element. $\qquad \blacksquare$
Thus, we can always assume that a maximal chain of prime ideals starts at a minimal prime and ends at a maximal ideal. But if $R$ is a noetherian local ring with principal maximal ideal $\mathfrak{m}$ and $\mathfrak{p}$ is a minimal prime of $R$, then $R / \mathfrak{p}$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$, and $\dim R = \sup_\mathfrak{p} \dim R / \mathfrak{p}$, as $\mathfrak{p}$ varies over the minimal primes.
Update. Georges Elencwajg pointed out in a comment that the first proof actually works without the assumption that $R$ is a domain, because $(1 - u t)$ is always invertible.
If $A$ is a ring (commutative with identity), $I$ an ideal of $A$, and $M$ a finitely generated $A$-module, then $M/IM$ is a finitely generated $(A/I)$-module. In fact, if $m_1,\ldots,m_r$ generate $M$ as an $A$-module, then there images in $M/IM$ generate it as an $(A/I)$-module. If $I$ is contained in the Jacobson radical of $A$ (meaning $I$ is contained in every maximal ideal of $A$) and $m_1+IM,\ldots,m_r+IM$ generate $M/IM$ as an $(A/I)$-module, then $m_1,\ldots,m_r$ generate $M$ as an $A$-module (assuming $M$ is finitely generated as an $A$-module). This is an application of a general form of Nakayama's lemma, which is also used to prove the special case where $A$ is local and $I=\mathfrak{m}$ is the unique maximal ideal (for $A$ local, its Jacobson radical is the maximal ideal).
Namely, let $N$ be the $A$-submodule of $M$ generated by $m_1,\ldots,m_r$. We then have $I(M/N)=M/N$. Indeed, given $m\in M$, by assumption, we have $m+IM=\sum_{i=1}^ra_im_i$ for some $a_i\in A$. Thus $m-\sum_{i=1}^ra_im_i\in IM$. This gives the non-trivial inclusion $M/N\subseteq I(M/N)$. By Nakayama, there is $a\in A$ with $a\equiv 1\pmod{I}$ such that $a(M/N)=0$. But because $I$ is contained in the Jacobson radical of $A$, $a$ must be a unit, and therefore $M=N$.
Best Answer
If you know that the dimension of a domain of finite type over $k$ is equal to the transcendence degree of its fraction field, then you know that $\dim(A/\mathfrak{p})=\mathrm{tr.deg}_k(K)$. Can you show using the definition of dimension that $\dim(A/\mathfrak{p})\leq\dim(A)$?