Existence of transcendence bases

Remarks

Apply Zorn's lemma to the algebraically independent sets $A$ with $S \subseteq A \subseteq T$, ordered by inclusion; a union of a chain is independent, since a polynomial relation involves only finitely many elements. Let $B$ be a maximal one. Each $t \in T$ is algebraic over $K(B)$, since otherwise $B \cup \set{t}$ would be a larger independent set. So $K(T)$ is algebraic over $K(B)$, and $L$ is algebraic over $K(B)$ by transitivity of algebraic extensions. For a finitely generated extension $L = K(t_1, \dots, t_n)$, Zorn's lemma is not needed, and one gets a transcendence basis among $t_1, \dots, t_n$.

Going deeper

For a student

This is the analogue of the theorem that every vector space has a basis, and the proof is the same Zorn's lemma argument. Together with the exchange property, it gives that all transcendence bases have the same cardinality, which defines the transcendence degree. It is used to write every extension as a purely transcendental extension followed by an algebraic one.