Extending a basis for a subspace of V to create a basis of V Can any set that forms a basis for a subspace of a vector space V be extended to form a basis for V?
 A: Yes! The argument is a simple Zorn's Lemma proof. Let $B$ be a given basis for a subspace $W$ of $V$. Use Zorn's Lemma to find a set $B'\subseteq V$ maximal with respect to the properties that $B\subseteq B'$ and that $B'$ is linearly independent. I claim $B'$ is a basis for $V$. If it weren't, let $x\in V$ be something not in the span of $B'$. Then $B'\cup\{x\}$ is linearly independent, contradicting maximality. 
A: Yes.  If your current set of linearly independent vectors does not span $V$ then choose any vector from $V$ that is not in the span and add it to your set of vectors.  You can (and should!) prove to yourself that because the vector you chose is not in the span of the other vectors in the set, the set remains linearly independent.
So until you find a basis (i.e. until the sets spans $V$) you will be able to keep adding vectors.  All you need to do now is prove that this process stops.  If you know $V$ is finite dimensional then you should know something about linearly independent sets having a maximal size.  In general, meaning if you don't know that fact or if $V$ is infinite dimensional, you can use Zorn's lemma to show that a basis exists.
Edit: As Cameron mentioned the axiom of choice, the place that occurs is when you use Zorn's lemma.  Zorn's lemma is equivalent to the axiom of choice.
A: Assuming that $V$ is a finite-dimensional vector space? Yes.
In general? It turns out that a linearly independent subset can be extended to a basis for the whole space if and only if the Axiom of Choice holds.
