Prove the existence of basis such that $L ⊆ B ⊆ S$ Suppose that in a vector space, $L$ there is a linearly independent set, $S$ is a set which spans the space and $L ⊆ S$, then there is a basis $B$ for the space such that $L ⊆ B ⊆ S$.
The proof is related to a standard proof of the weaker result:
Every vector space has a basis.
And a standard proof is: 

It says that we can use a similar way to prove the statement above in my question. So may I please ask how can I modify the argument? Thanks a lot! 
 A: Just define $\mathcal{L}$ to be the collection of all linearly independent subsets of $V$ which contain $L$ and are contained in $S$.  The rest of the proof (with very minor modifications) then shows that $\mathcal{L}$ has a maximal element $B$ (specifically, the modifications you have to make are that $L\in\mathcal{L}$ rather than $\emptyset\in\mathcal{L}$ and you must check that $\bigcup\mathcal{C}$ contains $L$ and is contained in $S$).
You then have to check that "the last lemma" referred to in the second sentence still shows that $B$ spans $V$.  I don't know the exact statement or proof you have for "the last lemma", but presumably it says something like if $B$ is linearly independent and $x$ is not in the span of $B$, then $B\cup\{x\}$ is linearly independent.  So in this case, if $B$ does not span $V$, its span cannot contain all of $S$ (since $S$ spans $V$), so you can choose $x$ to be an element of $S$.  Then $B\cup \{x\}$ will be an element of $\mathcal{L}$, contradicting maximality of $B$.
