# Subtle point in the proof that each finite spanning set of a vector space contains a basis

I saw the following proof of the fact that each finite spanning set $$S$$ of a vector space $$V$$ has a subset $$B\subseteq S$$, which is a basis of $$V$$.

Lemma. If $$T$$ is a minimal spanning set (i.e., $$T$$ is a spanning set and there is no spanning set which is a proper subset of $$T$$), then $$T$$ is a basis.

Proof. We have to show that $$T$$ is linearly independent. If not, then there is an $$x\in T$$ such that $$x$$ is a linear combination of $$T\setminus\{x\}$$. But then $$T\setminus\{x\}$$ is a spanning set which is a proper subset of $$T$$, contradicting the minimality of $$T$$.

Theorem. Let $$S$$ be a finite spanning set of $$V$$. Then there is a $$B\subseteq S$$ which is a basis of $$V$$.

Proof. Consider the following algorithm: if $$S$$ is linearly dependent, there is an $$x$$ such that $$x$$ is a linear combination of $$S\setminus \{x\}$$. In this case, set $$S:=S\setminus \{x\}$$ and repeat this step; else, stop.

At the end of the execution of the algorithm, $$S$$ is still a spanning set (because we only deleted redundant vectors). Also $$S$$ is then a minimal spanning set, because otherwise we would have deleted more vectors. Thus, by the lemma, $$S$$ is a basis.

My question: Is it really necessary to use the lemma? Doesn't the following argument suffice: Also $$S$$ is linearly independent, since otherwise, the algorithm wouldn't have ended.