= Solution
{c}
We use three theorems for an <integral extension> $S\subseteq R$, none requiring <Noetherian> hypotheses. The <Lying-over theorem> says that each <prime ideal> $\mathfrak p$ of $S$ is $\mathfrak q\cap S$ for some <prime ideal> $\mathfrak q$ of $R$. The <going-up theorem> says that if $\mathfrak p\subseteq\mathfrak p'$ in $S$ and $\mathfrak q\cap S=\mathfrak p$, there is $\mathfrak q'\supseteq\mathfrak q$ with $\mathfrak q'\cap S=\mathfrak p'$. The <incomparability theorem for integral extensions> says that comparable <prime ideals> of $R$ with the same contraction to $S$ are equal.
Take any strict chain
$$
\mathfrak q_0\subsetneq\cdots\subsetneq\mathfrak q_n
$$
in $R$. The <contraction of an ideal> operation gives $\mathfrak p_j=\mathfrak q_j\cap S$, which are <prime ideals> of $S$, by the proof in Question 1(b), and remain strictly increasing by the <incomparability theorem for integral extensions>. Thus every chain length in $R$ occurs in $S$, giving $\dim R\leq\dim S$.
Conversely, take any strict chain $\mathfrak p_0\subsetneq\cdots\subsetneq\mathfrak p_n$ in $S$. Use the <Lying-over theorem> to choose $\mathfrak q_0$ over $\mathfrak p_0$, then repeatedly use the <going-up theorem> to obtain $\mathfrak q_0\subseteq\cdots\subseteq\mathfrak q_n$ with the prescribed contractions. Each inclusion must be strict, since its contractions are distinct. Hence every chain length in $S$ occurs in $R$, giving $\dim S\leq\dim R$.
Taking suprema proves <integral extensions preserve Krull dimension>:
$$
\boxed{\dim R=\dim S.}
$$
The argument works equally well when the dimensions are infinite, since it compares all finite chain lengths.
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