# Containment of extension fields in a Bourbaki proof: $K(S)\subset K(B)$?

The theorem 3 in the section 14 (V.110) of Bourbaki's Algebra II states that

Let $E$ be an extension of a field $K$. All transcendence bases of $E$ over $K$ have the same cardinal.

I have a question about the proof in the case that a transcendence bases is infinite. In the following, $B$ and $B'$ are two transcendence bases of $E$ over $K$.

Every $x \in B$ is algebraic over $K(B')$ and so there exists a finite subset $S(x)$ of $B'$ such that $x$ is algebraic over $K(S(x))$. Write $S=\bigcup_{x \in B}S(x)$, then $S \subset B'$ and ...

... But since every element of $B$ is algebraic over $K(S)$ and $E$ is algebraic over $K(B)$, we conclude that $E$ is algebraic over $K(S)$ (V. p.19 Prop. 3).

The proposition referred to in the last line states that

Let $E$ and $F$ be two extension fields of a field $K$ such that $K\subset E\subset F$. For $F$ to be algebraic over $K$ it is necessary and sufficient that $E$ should be algebraic over $K$ and $F$ algebraic over $E$.

It seems to be assumed that $K(B)$ is an algebraic extension over $K(S)$. I see that $K(B)$ is algebraic over $K(S)$. But, how do we know that $K(S) \subset K(B)$ ?

We don't know that $K(S)\subset K(B)$, but we do have $K(S)\subset K(S\cup B)\subset E$, with $K(S\cup B)$ algebraic over $K(S)$ and $E$ algebraic over $K(S\cup B)$, so the proposition applies.