A subalgebra generated by finitely many integral elements is module-finite

Dispose of the zero base ring first. If $A=0$ then $1_A=0_A$, and a subring shares the identity and zero of the ambient ring, so $1_B=0_B$ and $B=0$; every subalgebra of $B$ is then the zero module over $A$, generated by the empty list. For the rest of the argument assume $A\mathrm{\ne }0$, so that $1_B\mathrm{\ne }{0}_B$ and every subring of $B$ containing $A$ is a nonzero commutative ring.

The case $n=0$: the subalgebra $A[]$ generated by the empty list is the smallest subring of $B$ containing $A$, which is $A$ itself, and $A$ is generated as an $A$-module by $1_A$.

Let $k\in \mathbb{N}$ and suppose $A_k:=A[b_1,\dots ,b_k]$ is module-finite over $A$.

Write $A_{k+1}:=A[b_1,\dots ,b_{k+1}]$. Both $A_{k+1}$ and $A_k[b_{k+1}]$ are the smallest subring of $B$ containing $A$ and $b_1,\dots ,b_{k+1}$, so they are equal. The element $b_{k+1}$ is a root of some monic $f\in A[X]$; the coefficients of $f$ lie in $A\subseteq A_k$, so $f$ is a monic polynomial in $A_k[X]$ with the same coefficients, and evaluating it at $b_{k+1}$ gives the same element of $B$, namely $0$. Hence $b_{k+1}$ is integral over $A_k$. Since $A_k$ is a nonzero commutative subring of $B$ by step 1.1, the integrality criterion gives that $A_k[b_{k+1}]=A_{k+1}$ is a finitely generated $A_k$-module; with the assumption of step 1.3 that $A_k$ is a finitely generated $A$-module, transitivity makes $A_{k+1}$ a finitely generated $A$-module.

The base case of step 1.2 and the passage of step 2.1 give, by induction on $n$, that $A[b_1,\dots ,b_n]$ is module-finite over $A$ for every $n\in \mathbb{N}$.