Let $S$ and $R$ be commutative rings with $1$. This is the usual form of the Going up theorem that one encounters in commutative algebra texts:
Let $S$ be integral over $R$, and suppose that we have a chain of prime ideals $P_1\subseteq\cdots\subseteq P_n$ of $R$, and a chain of prime ideals $Q_1\subseteq\cdots\subseteq Q_m$ of $S$ with $m<n$. If $Q_i$ lies over $P_i$ for $1\leq i\leq m$, then there are $n-m$ prime ideals $Q_{m+1},\dots,Q_n$ of $S$ such that $Q_1\subseteq\cdots\subseteq Q_{m+1}\subseteq\cdots\subseteq Q_n$ and $Q_i$ lies over $P_i$ for $1\leq i\leq n$.
Clearly, by induction this holds for countably infinite chains of $R$. Suppose now that $\mathfrak{P}=\{P_i\}_{i\in I}$ is an arbitrary chain of prime ideals of $R$. Is there a chain $\mathfrak{Q}=\{Q_i\}_{i\in I}$ of prime ideals of $S$ that lies over $\mathfrak{P}$ (i.e., $Q_i\cap R=P_i$ for all $i\in I$)?