Vector Space has Basis between Linearly Independent Set and Spanning Set

Theorem
Let $V$ be a vector space over a field $F$, let $L$ be a linearly independent subset of $V$, let $S$ be a set that spans $V$, and suppose that $L\subset S\subset V$. Then $V$ has a basis, $B$, such that $L\subset B\subset S$.

In particular, since $L$ can be taken to be any singleton of $V$, and $S$ can be taken to equal $V$, every vector space has a basis.

Proof
$\def\C{\mathscr C}$ $\def\I{\mathscr I}$ $\DeclareMathOperator{\span}{span}$ Let $\mathscr{I}$ be the set of linearly independent subsets of $S$ that contain $L$, ordered by inclusion. Note that $L\in\I$, so $\I\ne\varnothing$. Let $\C$ be a nest in $\I$, and let $C=\bigcup\C$. Then $C$ is linearly independent: Suppose that for some $v_{1},\ldots,v_{n}\in C$ and $r_1,r_2,\ldots,r_n\in F$ such that $r_1\ne 0$, $\sum_{k=1}^n r_k v_k = 0$. Then there are $C_1,C_2, \ldots, C_n\in\C$ such that $v_k\in C_k$ for each $k\in\{1,2,\ldots,n\}$. Since $\C$ is a nest, $C_1\cup C_2\cup\cdots\cup C_n$ must equal $C_k$ for some $k\in\{1,2,\ldots,n\}$. But then $C_k\in\C$ and $C_k$ is linearly dependent, a contradiction. Thus $C$ is linearly independent. By Zorn's lemma, $\I$ has a maximal element $M$ (one that is not contained in any other element). Since $M\in\I$, $M$ is linearly independent. All that remains is to show that $M$ spans $V$. Suppose, to the contrary, that there exists a $v\in V\setminus\span(M)$. Then, since $S$ spans $V$, there must be an element $s$ of $S$ such that $s\notin\span(M)$. Then $M\cup\{s\}$ is linearly independent, and $M\cup\{s\}\supsetneq M$, contradicting the maximality of $M$. Thus $M$ is a linearly independent subset of $V$ that spans $V$, and therefore a basis for $V$.