Theorem 2

Let E be an extension of a field K, S a subset of E such that E is algebraic over $K(S)$ and T a subset of S which is algebraically free over K; then there exists a transcendence basis B of E over K such that $T \subset B \subset S$.

For the set of algebraically free subsets of S, ordered by inclusion, is a set of finite character (E, III, p. 34) by Prop. 3. By Th. 1 of E, III, p. 35 it has a maximal element B containing T, and B is a transcendence basis of E over K, by Prop. 7.