Let me give a detailed account of Mahdi's argument.

First, we have to suppose (which is not totally clear in the question), that we consider only *countably many* primes $\mathfrak{p}_i$ of height $1$.

Second, we will suppose that $R/\mathfrak{m}$ is *uncountable* for every maximal ideal $\mathfrak{m}$ of $R$.

Then, $\dim(S^{-1}T)=1$.

Indeed, replacing $R$ by its localisation at a maximal ideal we can suppose that $R$ is in addition local. Then, in order to show that $\dim(S^{-1}T)=1$ we will show that every non-zero prime of $S^{-1}T$ has height $1$. So, let $\mathfrak{p}$ be a non-zero prime of $S^{-1}T$. Then, there exists a non-zero prime $\mathfrak{p}'$ of $T$ with $S\cap\mathfrak{p}'=\emptyset$ such that $\mathfrak{p}=S^{-1}\mathfrak{p}'$, and moreover $\mathfrak{p}$ and $\mathfrak{p}'$ have the same height. Thus it suffices to show that $\mathfrak{p}'$ has height $1$. The condition $S\cap\mathfrak{p}'=\emptyset$ is equivalent to $\mathfrak{p}'\subseteq\bigcup_{i\in\mathbb{N}}\mathfrak{q}_i$. Since $R$ is local with uncountable residue field, the same holds for $T$. Hence, countable prime avoidance holds in $T$, and so we see that there exists $i\in\mathbb{N}$ such that $\mathfrak{p}'\subseteq\mathfrak{q}_i$. As $\mathfrak{p}'$ is non-zero, $\mathfrak{q}_i$ has height $1$ and $T$ is a domain, it follows that $\mathfrak{p}'=\mathfrak{q}_i$ has height $1$ as desired.

(Using another variant of countable prime avoidance, also proven by Sharp and Vamos, gives the same conclusion under the hypothesis that $R$ is a complete local ring.)