Consider the localization $S_\mathfrak{p} = S \otimes_R R_\mathfrak{p}$. Since $S$ is integral over $R$, the extension $R_\mathfrak{p} \subseteq S_\mathfrak{p}$ is also integral (integrality is preserved under localization).

Claim: $S_\mathfrak{p} \neq 0$.

If $S_\mathfrak{p} = 0$, then $1 = 0$ in $S_\mathfrak{p}$, so there exists $s \in R \setminus \mathfrak{p}$ with $s \cdot 1 = 0$ in $S$. But $S$ is integral over $R$, hence faithful, contradicting $1 \neq 0$ in $S$.

Since $S_\mathfrak{p} \neq 0$ is a non-zero ring, it has a maximal ideal $\mathfrak{M}$. Let $\mathfrak{P} = \mathfrak{M} \cap S$.

Verification: We have: $\mathfrak{P} \cap R = (\mathfrak{M} \cap S) \cap R = \mathfrak{M} \cap R = \mathfrak{p} R_\mathfrak{p} \cap R = \mathfrak{p}$

The penultimate equality uses that $\mathfrak{M}$ is maximal over $\mathfrak{p} R_\mathfrak{p}$ in the integral extension $R_\mathfrak{p} \subseteq S_\mathfrak{p}$, so it contracts to $\mathfrak{p} R_\mathfrak{p}$.

Thus $\mathfrak{P}$ is a prime in $S$ lying over $\mathfrak{p}$.

($\Rightarrow$): Suppose $s$ satisfies $s^n + a_{n-1}s^{n-1} + \cdots + a_0 = 0$ with $a_i \in R$.

Then $s^n \in R[s^{n-1}, \ldots, s, 1]$, so: $R[s] = R \cdot 1 + R \cdot s + \cdots + R \cdot s^{n-1}$

is generated by $\{1, s, \ldots, s^{n-1}\}$ as an $R$-module, hence finitely generated.

($\Leftarrow$): Suppose $R[s]$ is finitely generated as an $R$-module, say by $\{m_1, \ldots, m_k\}$.

Since $m_i \in R[s]$, we can write $sm_i = \sum_{j=1}^k a_{ij} m_j$ for some $a_{ij} \in R$. In matrix form: $(sI - A)\mathbf{m} = 0$

where $A = (a_{ij})$ and $\mathbf{m} = (m_1, \ldots, m_k)^T$.

Multiplying by $\text{adj}(sI - A)$ (the adjugate matrix): $\det(sI - A) \mathbf{m} = 0$

Since $1 \in R[s]$ can be written as $1 = \sum r_i m_i$, we have: $\det(sI - A) = 0$

Expanding the determinant gives a monic polynomial equation for $s$: $s^k + c_{k-1}s^{k-1} + \cdots + c_0 = 0$

with coefficients $c_i \in R$. Thus $s$ is integral over $R$.

Let $R \subseteq S \subseteq T$ with $S$ integral over $R$ and $T$ integral over $S$. We show $T$ is integral over $R$.

Let $t \in T$. Since $t$ is integral over $S$: $t^n + s_{n-1}t^{n-1} + \cdots + s_0 = 0$

with $s_i \in S$. Let $R' = R[s_0, \ldots, s_{n-1}] \subseteq S$.

Since each $s_i$ is integral over $R$, we have $R[s_i]$ finitely generated as an $R$-module. Therefore: $R' = R[s_0, \ldots, s_{n-1}]$

is finitely generated as an $R$-module (finite extensions of finite extensions are finite).

Now $t$ satisfies a monic polynomial over $R'$, so $R'[t]$ is finitely generated as an $R'$-module. Since $R'$ is finitely generated as an $R$-module, we conclude: $R[t] \subseteq R'[t]$

is finitely generated as an $R$-module (composition of finite extensions).

By the module characterization, $t$ is integral over $R$.