I am trying to show the following statement:
Every module of finite type, i.e. finitely generated, has a minimal system of generators (a minimal system of generators $\mathcal S$ is a system of generators such that no proper subset of $\mathcal S$ generates the module).
I am not so sure if my idea is correct, so I'll tell what I've thought of, I would appreciate any corrections.
Let $M$ be a module of finite type. Then there exists a finite set $\mathcal S=\{x_1,x_2,...,x_n\}$ such that $M=<x_1,...,x_n>$. If $\mathcal S$ has no proper subset set that generates $M$ then we are done. Suppose $\mathcal S'$ is a proper subset of $\mathcal S$ that generates $M$. Then $\mathcal S'$ is of the form $\mathcal S'=\{x_{i_1},x_{i_2},...,x_{i_j}\}$ with $i_j<n$ and $x_{i_k} \in S$ for all $1\leq k \leq j$. If $\mathcal S'$ is minimal, we are done, if not, repeat the same process than before to extract a proper subset $\mathcal S'' \subset S'$ that generates $M$. If we could extract a minimal system before the $n-1$ step then we are done. If that wasn't the case, then we end up with a subset $\mathcal P \subset S$ with just one element. Then $P=\{x\}$ generates $M$ and it is clear that $P$ is minimal system. In any case, we could construct a minimal system extracted from $S$. From here it follows the statement.
Your proof is basically correct. But $\{x\}$ is not always minimal (look at $M=0$).
Here is a quick proof: We prove by induction on $n$ that every module generated by $n$ elements has a minimal generating set taken from these elements. The case $n=0$ is clear, the minimal generating set is $\emptyset$. Now let $x_1,\dotsc,x_n$ be a generating set. If it is minimal, we are done. Otherwise, we can erase some element, say $x_1$, so that $x_2,\dotsc,x_n$ is a generating set. By induction hypothesis, we can choose some of these elements which constitute a minimal generating set. QED