A module-finite algebra over a Noetherian ring is a Noetherian ring, and so is every ring between the two

Let $B^{\mathrm{\prime }}$ be any commutative $A$-algebra that is module-finite over $A$. Then $B^{\mathrm{\prime }}$ is a finitely generated module over the Noetherian ring $A$, hence a Noetherian $A$-module: every $A$-submodule of $B^{\mathrm{\prime }}$ is finitely generated over $A$. This applies in particular to $B^{\mathrm{\prime }}=B$.

First clause. Let $\mathfrak{b}$ be an ideal of $B^{\mathrm{\prime }}$. It is an additive subgroup of $B^{\mathrm{\prime }}$ closed under the $A$-action, since $a\cdot x={\eta }_{B^{\mathrm{\prime }}}(a)x\in \mathfrak{b}$ for $x\in \mathfrak{b}$, so it is an $A$-submodule and therefore generated over $A$ by finitely many $x_1,\dots ,x_k\in \mathfrak{b}$. Every element of $\mathfrak{b}$ is then ${\sum }_i{\eta }_{B^{\mathrm{\prime }}}(a_i)x_i$, which lies in the ideal $(x_1,\dots ,x_k)$ of $B^{\mathrm{\prime }}$; and that ideal is contained in $\mathfrak{b}$ because each $x_i$ lies in $\mathfrak{b}$. So $\mathfrak{b}=(x_1,\dots ,x_k)$ is finitely generated as an ideal, and $B^{\mathrm{\prime }}$ is a Noetherian ring. Taking $B^{\mathrm{\prime }}=B$ gives the first clause.

Second clause, first half. Suppose the $A$-algebra structure map is the inclusion $A\subseteq B$ and $C$ is a subring of $B$ with $A\subseteq C\subseteq B$. Then $C$ is an additive subgroup of $B$ closed under the $A$-action, because $a\cdot x=ax$ is a product of two elements of $C$; so $C$ is an $A$-submodule of the Noetherian $A$-module $B$ and is therefore finitely generated as an $A$-module, that is, module-finite over $A$.

Third clause. Let $M$ be a finitely generated $B$-module, say generated over $B$ by $u_1,\dots ,u_n$, and let $b_1,\dots ,b_m$ generate $B$ over $A$. The $mn$ products $b_i{u}_j$ generate $M$ as an $A$-module, so $M$ is a finitely generated $A$-module and hence a Noetherian $A$-module.

Step 2.2 makes $C$ a commutative $A$-algebra, through the inclusion $A\subseteq C$, that is module-finite over $A$; steps 1.1 and 2.1 were proved for an arbitrary such algebra, so applying them with $B^{\mathrm{\prime }}=C$ makes $C$ a Noetherian ring. With steps 2.1 and 2.3 this establishes all three clauses.