I borrowed the idea of the Bourbaki's proof of Krull-Akizuki theorem.
Lemma 1
Let A be a weakly Artinian integral domain.
Let $M$ be a torsion $A$-module of finite type.
Then $leng_A M$ is finite.
Proof:
Let $x_1, ..., x_n$ be generating elements of $M$.
There exists a non-zero element $f$ of $A$ such that $fx_i = 0$, $i = 1, ..., n$.
Let $\psi:A^n \rightarrow M$ be the morphism defined by $\psi(e_i) = x_i$, $i = 1, ..., n$,
where $e_1, ..., e_n$ is the canonical basis of $A^n$.
Since $leng_A A^n/fA^n$ is finite and $\psi$ induces a surjective mophism $A^n/fA^n \rightarrow M$, $leng_A M$ is finite.
QED
Lemma 2
Let A be a weakly Artinian integral domain.
Let $K$ be the field of fractions of $A$.
Let $M$ be a torsion-free $A$-module of finite type.
Let $r = dim_K M \otimes_A K$.
Let $f$ be a non-zero element of $A$.
Then $leng_A M/fM \leq r(leng_A A/fA)$
Proof:
There exists a $A$-submodule $L$ of $M$ such that $L$ is isomorphic to $A^r$ and $Q = M/L$ is a torsion module of finite type over $A$.
Hence, by Lemma 1, $leng_A Q$ is finite.
Let $n \geq 1$ be any integer.
The kernel of $M/f^nM \rightarrow Q/f^nQ$ is $(L + f^nM)/f^nM$ which is isomorphic to $L/(f^nM \cap L)$.
Since $f^nL \subset f^nM \cap L$,
$leng_A M/f^nM \leq leng_A L/f^nL + leng_A Q/f^nQ \leq leng_A L/f^nL + leng_A Q$.
Since $M$ is torsion-free, $f$ induces isomorphism $M/fM \rightarrow fM/f^2M$.
Hence $leng_A M/f^nM = n(leng_A M/fM)$.
Similarly $leng_A L/f^nL = n(leng_A L/fL)$.
Hence $leng_A M/fM \leq leng_A L/fL + (1/n) leng_A Q$.
Since $L$ is isomorphic to $A^r$, $leng_A L/fL = r(leng_A A/fA)$. Hence, by letting $n \rightarrow \infty$, $leng_A M/fM \leq r(Leng_A A/fA)$.
QED
Lemma 3
Let A be a weakly Artinian integral domain.
Let $K$ be the field of fractions of $A$.
Let $M$ be a torsion-free $A$-module.
Suppose $r = dim_K M \otimes_A K$ is finite.
Let $f$ be a non-zero element of $A$.
Then $leng_A M/fM \leq r(Leng_A A/fA)$
Proof:
Let $(M_i)_I$ be the family of finitely generated $A$-submodules of $M$.
$M/fM = \cup_i (M_i + fM)/fM =\cup_i M_i/(M_i \cap fM)$.
Since $fM_i \subset M_i \cap fM$, $M_i/(M_i \cap fM)$ is isomorphic to a quotient of $M_i/fM_i$.
Hence, by Lemma 2, $leng_A M_i/(M_i \cap fM) \leq r(leng_A A/fA)$.
Hence, by By Lemma 4 of this, $leng_A M/fM \leq r(leng_A A/fA)$
QED
Lemma 4
Let A be a weakly Artinian integral domain.
Let $K$ be the field of fractions of $A$.
Let $L$ be a finite extension field of $K$.
Let $B$ be a subring of $L$ containing $A$.
Then $leng_A B/fB$ is finite for every non-zero element $f \in B$.
Proof:
Since $L$ is a finite extension of $K$, $a_rf^r + ... + a_1f + a_0 = 0$, where $a_i \in A, a_0 \neq 0$.
Then $a_0 \in fB$.
Since $B \otimes_A K \subset L$, $dim_K B \otimes_A K \leq [L : K]$.
Hence, by Lemma 3, $leng_A B/a_0B$ is finite.
Hence $leng_A B/fB$ is finite.
QED
Proof of the title theorem
By Lemma 2 of my answer to this, $B$ is weakly Artinian.
Hence, by this, we are done.
QED
