Finite generating sets and finitely-generated modules (All my rings are commutative with $1$.)
Reworded a little, a question in a previous Commutative Algebra exam goes like this:

Let
  
  
*
  
*$A$ denote a ring,
  
*$X$ denote an $A$-module
  
*$F$ denote a finite generating subset of $X$, and
  
*$\Lambda$ denote a (not-necessarily finite) generating subset of $X$.
  
  
  Show that $\Lambda$ includes a finite generating subset of $X$.

The obvious approach is to try to inductively choosing elements of $\Lambda$ in a way that is guided by $F$ somehow, until we've built a finite subset that generates all of $\Lambda$ and hence of all of $X$. I really have no idea how to do this though.
Ideas, anyone?
Answers phrased in the language of finitary closure operators and/or algebraic posets are especially welcome!
 A: Each element of $F$ can be written as a finite linear-combination of elements of $\Lambda$. Since there are finitely many elements in $F$, this gives rise to a finite subset $\Omega \subset \Lambda$ with $A\langle \Omega \rangle \supset A \langle F \rangle = X$.
A: Okay, so I've taken the answer MooS gave and generalized it a lot, by formulating everything in the language of closure systems. The essence of the proof is the same, however. Also, its a little wordy because I wanted to be 100% rigorous and precise. But fundamentally, the argument is the same.
For the remainder of this answer, let $X$ denote a closure system.

Definition 0. A subset $A \subseteq X$ is algebraic iff $A = \mathop{\bigcup}_{F \in \mathcal{P}_{\mathrm{fin}}(A)}\mathrm{cl}(F).$
Proposition 0. For all $A \subseteq X$, the following are equivalent:
  
  
*
  
*$A$ is algebraic
  
*$\mathop{\forall}_{x \in X} \left(x \in \mathrm{cl}(A) \rightarrow \mathop{\exists}_{F \in \mathcal{P}_{\mathrm{fin}}(A)}x \in \mathrm{cl}(F)\right)$

Proof. The following are equivalent.


*

*$A = \mathop{\bigcup}_{F \in \mathcal{P}_{\mathrm{fin}}(A)}\mathrm{cl}(F)$

*$A \subseteq \mathop{\bigcup}_{F \in \mathcal{P}_{\mathrm{fin}}(A)}\mathrm{cl}(F)$

*$\mathop{\forall}_{x \in X} \left(x \in \mathrm{cl}(A) \rightarrow x \in \mathop{\bigcup}_{F \in \mathcal{P}_{\mathrm{fin}}(A)}\mathrm{cl}(F)\right)$

*$\mathop{\forall}_{x \in X} \left(x \in \mathrm{cl}(A) \rightarrow \mathop{\exists}_{F \in \mathcal{P}_{\mathrm{fin}}(A)}x \in \mathrm{cl}(F)\right)$

Proposition 1. Let $\Lambda$ denote an algebraic generating subset of $X$. 
Then for all finitely-generated $G \subseteq X$, there exists finite $\Lambda_{\mathrm{fin}} \subseteq \Lambda$ such that $\mathrm{cl}(\Lambda_{\mathrm{fin}}) \supseteq G$.

Proof.
Since $\Lambda$ is algebraic, hence by Proposition 0, we have
$$\mathop{\forall}_{x \in X} \left(x \in \mathrm{cl}(\Lambda) \rightarrow \mathop{\exists}_{F \in \mathcal{P}_{\mathrm{fin}}(\Lambda)}x \in \mathrm{cl}(F)\right)$$
Therefore, since $\Lambda$ is a generating set, we have: $$\mathop{\forall}_{x \in X} \mathop{\exists}_{F \in \mathcal{P}_{\mathrm{fin}}(\Lambda)}x \in \mathrm{cl}(F)$$
Hence letting $G_{\mathrm{fin}}$ denote a finite generating set of $G$, as a special case, we get: $$\mathop{\forall}_{g \in G_{\mathrm{fin}}} \mathop{\exists}_{F \in \mathcal{P}_{\mathrm{fin}}(\Lambda)}g \in \mathrm{cl}(F)$$
Using the principle of finite choice, this can be rewritten:
$$\mathop{\exists}_{F : G \rightarrow \mathcal{P}_{\mathrm{fin}}(\Lambda)}\mathop{\forall}_{g \in G_{\mathrm{fin}}} g \in \mathrm{cl}(F(g))$$
So let $F : G_{\mathrm{fin}} \rightarrow \mathcal{P}_{\mathrm{fin}}(\Lambda)$ denote any such function. Define:
$$\Lambda_{\mathrm{fin}} = \bigcup_{g \in G_{\mathrm{fin}}} F(g)$$
Since $G_{\mathrm{fin}}$ is finite, hence $\Lambda_{\mathrm{fin}}$ is finite.
We need to show that $\mathrm{cl}(\Lambda_{\mathrm{fin}}) \supseteq G$.
It suffices to show that $\mathrm{cl}(\Lambda_{\mathrm{fin}}) \supseteq G_{\mathrm{fin}}$.
We know that
$$\mathop{\forall}_{g \in G_{\mathrm{fin}}} g \in \mathrm{cl}(F(g))$$
In other words:
$$G_{\mathrm{fin}} \subseteq \bigcup_{g \in G} \mathrm{cl}(F(g)).$$
So
$$G_{\mathrm{fin}} \subseteq \mathrm{cl}\left(\bigcup_{g \in G} F(g)\right).$$
In other words:
$$G_{\mathrm{fin}} \subseteq \mathrm{cl}(\mathrm{\Lambda}_{\mathrm{fin}}).$$
as required.

Corollary. Let $\Lambda$ denote an algebraic generating subset of $X$. 
If $X$ is finitely-generated, then $\Lambda$ includes a finite generating set of $X$.

