Assume that $A \ \cup \ \{v\}$ is not independent. We know that $A$ is independent, so any finite linear combination of elements contained within $A$ can only equal $0$ in the trivial case. For $A \ \cup \ \{v\}$ to not be independent, it would thus have to be true that a finite linear combination of elements in $A$, along with $v$ must be equal to $0$ in a non-trivial case: $$c v \ + \ \displaystyle\sum_{i \ = \ 1}^{N} \ c_i a_i \ = \ 0$$ where each $a_i \ \in \ A$. It also must be true that $c \ \neq \ 0$, as if it were, then this would simply be a non-trivial linear combination of elements of $A$, contradicing the fact that $A$ is independent. We can thus write: $$v \ = \ - \displaystyle\sum_{i \ = \ 1}^{N} \frac{c_i}{c} a_i$$ Thus, $v$ is a linear combination of elements of $A$, and is thus in the span of $A$. This is a contradiction, implying that $A \ \cup \ \{v\}$ must be independent.

Let us consider the set $\mathcal{A}$ of all independent sets in $V$. Let us also consider some subset $A$ of $\mathcal{A}$ that is simply ordered by proper inclusion. We wish to prove that any such subset will have an upper bound contained within $\mathcal{A}$. It would follow (by Zorn's lemma) that for every element of the subset, there always exists another which contains it as a proper subset.

Assume that the union of all elements of $A$, which we denote by $\bigcup A$ is not independent. This would imply that there exists some $\textbf{v} \ \in \ \bigcup A$ such that: $$\textbf{v} \ = \ \displaystyle\sum_{n \ = \ 1}^{N} c_n v_n$$ with each $v_n \ \in \ \bigcup A$. By definition, $\textbf{v}$ and each $v_n$ must be contained in at least one element of $A$. It follows from the finite axiom of choice that we can pick such elements of $A$, which we denote as: $$A_{V} \ = \ \{A_{\textbf{v}}, A_{v_1}, \ ..., \ A_{v_n}\}$$ Since $A$ is order by proper inclusion, it follows that $A_{V}$ is also ordered by proper inclusion. This implies that there exists a maximal element $A_{v_i}$ of $A_V$ (since $A_V$ is finite). Since every element of $A_V$ is contained in $A_{v_i}$, it follows that: $$A_{v_i} \ = \ \bigcup A_{V}$$ We know that $A_{v_i}$ is independent, so it follows that $\bigcup A_V$ is independent. However, the element $\textbf{v}$ of $\bigcup A_V$ is equal to a finite linear combination of other elements of $A_V$, which implies that $A_V$ is not independent.

This is a clear contradiction, which implies that $\bigcup A$ is independent. It follows that $\bigcup A$ is in $\mathcal{A}$ and is an upper bound for $A$. This implies that $\mathcal{A}$ has a maximal element, by Zorn's lemma.

We have proven that $\mathcal{A}$ has a maximal element, which we will call $A$. By the first lemma, it must be true that $A \ \cup \ \{v\} \ \in \ \mathcal{A}$ for some $v \ \in \ V$ that does not belong to the span of $V$. If such an element existed, this would contradict the fact that $A$ is a maximal element. Thus, it must be true that there exists no $v \ \in \ V$ that is not in the span of $A$. Thus, $A$ is independent and spans the vector space. By definition, $A$ is a basis of $V$.