We use the definitions of my answer to this question.
Definition 1
Let $A$ be a commutative ring.
Let $B$ be a commutative $A$-algebra.
Suppose $leng_A B$ is finite.
Then we say $B$ is an Artinian $A$-algebra.
Lemma 1
Let $A$ be a commutative ring.
Let $B$ be an Artinian $A$-algebra.
Let $\Lambda$ be nonempty set of ideals of $B$.
Then there exist a maximal element and a minimal element in $\Lambda$.
Proof:
We note that every ideal of $B$ can be regarded canonically as an A-module.
Let $r = sup$ {$leng_A I; I \in \Lambda$}.
Since $r$ is finite, there exists $I \in \Lambda$ such that $r = leng$ $I$.
By Lemma 4 of my answer to this, $I$ is a maximal element of $\Lambda$.
The existence of a minimal element is proved similarly.
QED
Lemma 2
Let $A$ be a commutative ring.
Let $B$ be an Artinian $A$-algebra.
Then $leng_B B$ is finite, namely $B$ is an Artinian ring as defined in Definition 1 in this.
Proof:
Let $\Lambda$ be the set of ideals $I$ of $B$ such that $leng_B I$ is finite.
Since $0 \in \Lambda$, $\Lambda$ is not empty.
By Lemma 1, there exists a maximal element $I \in \Lambda$.
Suppose $I \neq B$.
Then, by Lemma 1, there exists an ideal $J$ of $B$ such that $I \subset J$ and $J/I$ is a simple $B$-module.
Since $leng_B J = leng_B I + 1$, $J \in \Lambda$.
This is a contradiction.
Hence $B = I$.
QED
Definition 2
Let $A$ be a commutative ring.
Let $B$ be a commutative $A$-algebra.
Suppose $leng_A B/fB$ is finite of for every non-zero element $f \in B$.
Then we say $B$ is a weakly Artinian $A$-algebra.
By Lemma 2, $B$ is a weakly Artinian ring.
Proof of the title theorem
By Lemma 2, $B$ is a weakly Artinian ring.
Hence the assertions of the title theorem follow immediately from Lemma 2 and this.
QED
