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