I don't know where to start on this proof. I know that a linearly independent set is a set where $c_1v_1+c_2v_2+\cdots+c_nv_n = 0$ has no non-trivial solutions other than $c_1=c_2=...=c_n=0$. I know that a basis is a linearly independent set and a spanning set for the vector space.
How can I use that to figure out this proof?
On
Let $$ \mathcal Z=\{\,A\subseteq V\mid S\subseteq A, A\text{ is linearly independent}\,\}$$ endowed with $\subseteq $ as partial order. If $\{\,A_i\mid i\in I\,\}\subseteq \mathcal Z$ with $A_i\subseteq A_j\lor A_j\subseteq A_i$ for all $i,j\in I$, then $A:=S\cup \bigcup_{i\in I}A_i$ is an elemtn of $\mathcal Z$ with $A_i\subseteq A$ for all $i\in I$. These are precisely the conditins that allow Zorn's lemma to be invoked. Thus let $B$ be a maximal element of $\mathcal Z$. Clearly, $B$ is linearly independent. And if $v\in V$, then either $v\in B$ (and so $v\in \operatorname{span}(B)$) or $B\cup\{v\}$ is not in $\mathcal Z$, i.e., tehre eixts a linear dependence among $B\cup\{v\}$; as this dependencemust involve $v$, we find that $v\in\operatorname{span}(B)$ again. Hence $B$ is a basis of $V$.
On
An interesting tool in proving theorems about bases and dimension is the “Steinitz exchange lemma”
If $A=\{v_1,\dots,v_m\}$ is a linearly independent set in $V$ and $B=\{w_1,\dots,w_n\}$ is a spanning set for $V$, then $m\le n$ and there is a permutation $\sigma$ of $\{1,2,\dots,n\}$ such that $\{v_1,\dots,v_m,w_{\sigma(m+1)},\dots,w_{\sigma(n)}\}$ is a spanning set of $V$.
Proof. By induction on $m$. If $m=0$ there is nothing to prove. Suppose we have substituted $k$ elements of $B$ with the first $k$ elements of $A$ and $k<m$ in the sense we have found a permutation $\sigma$ of $\{1,\dots,n\}$ so that $$ \{v_1,\dots,v_k,w_{\sigma(k+1)},\dots,w_{\sigma(n)}\} $$ is a spanning set for $V$. Also the set $\{v_1,\dots,v_k,v_{k+1},w_{\sigma(k+1)},\dots,w_{\sigma(n)}\}$ is a spanning set and, moreover $$ w_{k+1}=\alpha_1w_1+\dots+\alpha_kw_k+ \alpha_{k+1}w_{\sigma(k+1)}+\dots+\alpha_nw_{\sigma(n)} $$ Now, at least one among $\alpha_{k+1},\dots,\alpha_n$ is nonzero, otherwise $w_{k+1}$ would belong to the span of $\{w_1,\dots,w_k\}$, which is impossible because $A$ is linearly independent. Let $h>k$ be minimal such that $\alpha_h\ne0$; compose $\sigma$ with the transposition of $k+1$ and $h$, call the composition $\tau$. Thus, by a simple algebraic manipulation, $$ w_{\tau(k+1)}=\beta_1v_1+\dots+\beta_kv_k+\beta_{k+1}v_{k+1} +\beta_{k+2}w_{\tau(k+2)}+\dots+\beta_{n}w_{\tau(n)} $$ for suitable scalars $\beta_1,\dots,\beta_n$ and so we conclude that $$ \{v_1,\dots,v_{k+1},w_{\tau(k+2)},\dots,w_{\tau(n)}\} $$ is a spanning set for $V$. QED
Now take $B$ a basis for $V$.
If $S$ is already a spanning set, then it is a basis, and we're done. If it is not, there is there some vector $w\in V\setminus S$. If you form the set $S'=S\cup \{w\}$, you should be able to show that it is linearly independent. Also, its span strictly contains the span of $S$.
If $S'$ is still not a basis, find a vector $w_2$ outside of its span, and form $S''$. Et cetera.
How do we know that this process will eventually span the vector space? We know that $V$ is finite-dimensional; suppose $\dim V=n$. As soon as we have $n$ vectors is our enlarged-$S$, it will have to be a basis. You should work out why that claim is true.