Let $A \subset B$ be two rings. I know that an element $x \in B$ is integral over $A$ iff $A[x]$ is contained in a finitely generated $A$-module $T \subset B$. I also know that if $b_1,...,b_n$ are integral over $A$ then $A[b_1,...,b_n]$ is a finitely generated $A$-module.
Why does these imply that the set of elements of $B$ that are integral over $A$ form a ring?
Let $a,b$ be two elements of $B$ integral over A. Your second assertion implies that $a\in{}M, b\in{}N$ which are both finitelly generated submodules of $B$. Then $MN=\{\sum{}a_ib_j, a_i\in{}M,b_j\in{}N\}$ is a finitelly generated submodule of $B$. Notice that $MN$ contains both $a+b$ and $ab$ and the conclusion follows...
Note that you are taking the $x$-subalgebra generated by $x$ (I mean $A[x]$) but you really need a finitely generated submodule of $A$ (let's say $M$) such that $xM\subset{}M$. Since Dedekind's proof uses this inclusion to work with Cramer's rule it is this essential property that's needed.