Let $K$ be a field and $E$ a $K$-vector space. Let $S$ be a generating system $E$ and $L$ a free subset of $E$ such that $L\subset S$. Then there exists a basis of $E$ such that $L\subset B\subset S$.
Let $\mathcal{L}$ be the set of free subsets of $E$ contained in $S$. This set is inductive. Then the set $\mathcal{M}$ of free subsets of $E$ containing $L$ and contained in $S$ is also inductive. By Zorn's lemma, there exists a maximal $B$ in $\mathcal{M}$. We want to show that $E\subset\text{span}(B)$. Suppose not: let $y\in E$ such that $y\not\in\text{span}(B)$. Then $B\cup\{y\}$ is a free subset of $E$ and $L\subset B\cup\{y\}$. I still need to show that $B\cup\{y\}\subset S$ in order to get a contradiction. But why should $y\in S$?
It doesn't have to be true. But instead of taking some $y\in E$ such that $y\notin\operatorname{span}(B)$, take $y\in S$ such that $y\notin\operatorname{span}(B)$. Such a $y$ must exist, because otherwise $S\subset\operatorname{span}(B)$, and therefore $\operatorname{span}(B)$ would be the whole space.