There's just one step in this proof I can't see for the life of me.
Set up: We have a field K and an algebraically closed field $\Omega$. $(B, g)$ is maximal in the set $\Sigma$ of ordered pairs $(A, f)$ where $A$ is a subring of K and $f$ a homomorphism into $\Omega$, where $\Sigma$ has the partial order $(A, f) \leq (A', f')$ if $A$ is a subring of $A'$ and $f'|_{A} = f$. The overall claim is that $(B, g)$ is a valuation of $K$. We let $M$ be the unique maximal ideal of $B$ (which exists). We take $x \in K$ with $x \neq 0$ and may assume that $M[x]$ is not the unit ideal of $B' = B[x]$(by a lemma) and so is contained in some maximal ideal $M'$. Let $k = B/M$ and $k' = B'/M'$.
The claim I don't understand: Since $k' = k[\bar{x}]$ for $\bar{x}$ the image of x in k' (which I see), $\bar{x}$ is algebraic over k.