Proof of Basis Extension Theorem 
Let $V$ be a finite dimensional vector space and $S=\{a_2,a_3,\ldots,a_n\}$, a linearly independent subset of V. Then either $S$ itself is a basis of $V$ or $S$ can be extended to form a basis of $V.$

Please prove the above.
 A: Suppose $I$ is an independent subset of $U$. If span $\{I\}=U$, then $I$ is already a basis of $U$. If span$\{I\}\neq U$, choose $u_1\in U$ such that $u_1 \notin $span$\{$I$\}$. Hence the set $I \cup \{u_1\}$ is independent by the Independent Lemma. 
(Recall the Independent Lemma: Let $\{v_1,v_2,\ldots,v_k\}$ be an independent set of vectors in a vector space $V$. If $u \in V$ but $u \notin$span$\{v_1,v_2,\ldots,v_k\},$then $\{u,v_1,v_2,\ldots,v_k\}$ is also independent.)
If span$\{I\cup\{u_1\}\}=U,$ we are done; otherwise choose $u_2\in U$ such that $u_2 \notin$span$\{I\cup \{u_1\}\}.$ Hence, $\{I\cup \{u_1,u_2\}\}$ is independent and the process continues. A basis of $U$ will be reached eventually. Why? This is because if no basis of $U$ is ever reached, then the process will create arbitrarily large independent sets in $V$, but this is impossible by the fact that because $V$ is finite dimensional and is therefore spanned by a finite set of vectors.
$Q.E.D.$ 
A: Hints: Show that
$$\begin{align*}\bullet&\;\;\;x\in\text{Span}\,\{a_2,...,a_n\}\iff \{a_2,...,a_n,x\}\;\;\text{is linearly $\bf\text{dependent}$}\\{}\\
\bullet&\;\;\;\dim V<\infty\implies\;\text{there exists a finite generating set}\end{align*}$$
