Let $V$ be a vector space (over a field), and let $\mathcal{S} \subset \mathcal{P}(V)$ be the set of all sets of linearly independent vectors, partially ordered by set inclusion.

Any chain $\mathcal{C} \subseteq \mathcal{S}$ is bounded above by $S_{\text{max}} = \bigcup_{S \in \mathcal{C}} S$ which is a linearly independent set of vectors because any finite subset of $S_{\text{max}}$ must be contained in some element of $\mathcal{C}$. Thus, by Zorn’s Lemma there exists at least one maximal element $M \in \mathcal{S}$.

Suppose for contradiction that $M$ does not span. Then there is some $v \in V$ which cannot be expressed as a finite linear combination of elements of $M$. But this implies $M' = M \cup \{v\} \in \mathcal{S}$ and $M \subset M'$, contradicting the maximality of M with respect to inclusion.

So $M$ must span, and since it is a linearly independent set, this makes it a basis.