Vector Space has Basis
Let $K$ be a division ring.
Let $V$ be a vector space over $K$.
Then $V$ has a basis.
The result follows from Vector Space has Basis between Linearly Independent Set and Spanning Set.
$S$ can be taken to be $V$, since $V$ trivially spans itself.
Therefore, $L$ and $S$ exist and $L \subseteq S$ so $V$ has a basis $B$ with $L \subseteq B \subseteq S$.