I borrowed the idea of the Bourbaki's proof of Krull-Akizuki theorem.
Definition Let $A$ be a not-necessarily commutative ring. Let $M$ be a left $A$-module. Suppose $M$ has a composition series, the lengths of each series are the same by Jordan-Hoelder theorem. We denote it by $leng_A M$. If $M$ does not have a composition series, we define $leng_A M = \infty$.
Lemma 1 Let $A = k[X]$ be a polynomial ring of one variable over a field $k$. Let $f$ be a non-zero element of $A$. Then $A/fA$ is a finite $k$-module.
Proof: Clear.
Lemma 2 Let $A = k[X]$ be a polynomial ring of one variable over a field $k$. Let $M$ be a torsion $A$-module of finite type. Then $M$ is a finite $k$-module.
Proof: Let $x_1, ..., x_n$ be generating elements of $M$. There exists a non-zero element $f$ of $A$ such that $fx_i = 0$, $i = 1, ..., n$. Let $\psi:A^n \rightarrow M$ be the morphism defined by $\psi(e_i) = x_i$, $i = 1, ..., n$, where $e_1, ..., e_n$ is the canonical basis of $A^n$. By Lemma 1, $A^n/fA^n$ is a finite $k$-module. Since $\psi$ induces a surjective mophism $A^n/fA^n \rightarrow M$, $M$ is a finite $k$-module. QED
Lemma 3 Let $A = k[X]$ be a polynomial ring of one variable over a field $k$. Let $M$ be an $A$-module. Then $length_A M < \infty$ if and only if $M$ is a finite $k$-module.
Proof: Suppose $length_A M < \infty$. Let $M = M_0 \supset M_1 \supset ... \supset M_n = 0$ be a composition series. Each $M_i/M_{i+1}$ is isomorphic to $A/f_iA$, where $f_i$ is an irreducible polynomial in $A$. Since $dim_k A/f_iA$ is finite by Lemma 1, $dim_k M$ is finite.
The converse is clear. QED
Lemma 4 Let $A$ be a not necessarily commutative ring. Let $M$ be a left $A$-module. Let $(M_i)_I$ be a family of $A$-submodules of $M$ indexed be a set $I$. Suppose $(M_i)_I$ satisfies the following condition.
$M = \cup_i M_i$, and for any $i, j \in I$, there exists $k \in I$ such that $M_i \subset M_k$ and $M_j \subset M_k$.
Then $leng_A M = sup_i leng_A M_i$.
Proof: Suppose $sup_i leng_A M_i = \infty$. Since $sup_i leng_A M_i \leq leng_A M$, $leng_A M = \infty$. Hence we can assume that $sup_i leng_A M_i = n < \infty$. Let $n = leng_A M_{i_0}$. For each $i \in I$, there exists $k \in I$ such that $M_{i_0} \subset M_k$ and $M_i \subset M_k$. Since $leng_A M_k = n$, $M_{i_0} = M_k$, $M_i \subset M_{i_0}$. Since $M = \cup_i M_i$, $M = M_{i_0}$. Hence $leng_A M = n$. QED
Lemma 5 Let $A = k[X]$ be a polynomial ring of one variable over a field $k$. Let $K$ be the field of fractions of $A$. Let $M$ be a torsion-free $A$-module of finite type. Let $r = dim_K M \otimes_A K$ Let $f$ be a non-zero element of $A$. Then $leng_A M/fM \leq r(leng_A A/fA)$
Proof: There exists a $A$-submodule $L$ of $M$ such that $L$ is isomorphic to $A^r$ and $Q = M/L$ is a torsion module of finite type over $A$. Hence, by Lemma 2, $Q$ is a finite $k$-module. The kernel of $M/f^nM \rightarrow Q/f^nQ$ is $(L + f^nM)/f^nM$ which is isomorphic to $L/(f^nM \cap L)$. Since $f^nL \subset f^nM \cap L$, $leng_A M/f^nM \leq leng_A L/f^nL + leng_A Q/f^nQ \leq leng_A L/f^nL + leng_A Q$. Since $M$ is torsion-free, $f$ induces isomorphism $M/fM \rightarrow fM/f^2M$. Hence $leng_A M/f^nM = n(leng_A M/fM)$. Similarly $leng_A L/f^nL = n(leng_A L/fL)$. Hence $leng_A M/fM \leq leng_A L/fL + (1/n) leng_A Q$. Since $L$ is isomorphic to $A^r$, $leng_A L/fL = r(leng_A A/fA)$. Hence $leng_A M/fM \leq r(Leng_A A/fA)$. QED
Lemma 6 Let $A = k[X]$ be the polynomial ring of one variable over a field $k$. Let $K$ be the field of fractions of $A$. Let $M$ be a torsion-free $A$-module. Suppose $r = dim_K M \otimes_A K$ is finite. Let $f$ be a non-zero element of $A$. Then $leng_A M/fM \leq r(Leng_A A/fA)$
Proof: Let $(M_i)_I$ be the family of finitely generated $A$-submodules of $M$. $M/fM = \cup_i (M_i + fM)/fM =\cup_i M_i/(M_i \cap fM)$. Since $fM_i \subset M_i \cap fM$, $M_i/(M_i \cap fM)$ is isomorphic to a quotient of $M_i/fM_i$. Hence, by Lemma 5, $leng_A M_i/(M_i \cap fM) \leq r(leng_A A/fA)$. Hence, by Lemma 4, $leng_A M/fM \leq r(leng_A A/fA)$ QED
Lemma 7 Let $A = k[X]$ be a polynomial ring of one variable over a field $k$. Let $K$ be the field of fractions of $A$. Let $L$ be a finite extension field of $K$. Let $B$ be a subring of $L$ containing $A$. Then $B/fB$ is a finite $k$-module for every non-zero element $f \in B$.
Proof: Since $L$ is a finite extension of $K$, $a_rf^r + ... + a_1f + a_0 = 0$, where $a_i \in A, a_0 \neq 0$. Then $a_0 \in fB$. Since $B \otimes_A K \subset L$, $dim_K B \otimes_A K \leq [L : K]$. Hence, by Lemma 6, $leng_A B/a_0B$ is finite. Hence $leng_A B/fB$ is finite. Hence, by Lemma 3, the assertion follows. QED
Lemma 8 Let $A$ be an integrally closed domain containing a field $k$ as a subring. Suppose $A/fA$ is a finite $k$-module for every non-zero element $f \in A$. Let $S$ be a multiplicative subset of $A$. Let $A_S$ be the localization with respect to $S$. Then $A_S$ is an integrally closed domain containing a field $k$ as a subring and $A_S/fA_S$ is a finite $k$-module for every non-zero element $f \in A_S$.
Proof: Let $K$ be the field of fractions of $A$. Suppose that $x \in K$ is integral over $A_S$. $x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 = 0$, where $a_i \in A_S$. Hence there exists $s \in S$ such that $sx$ is integral over $A$. Since $A$ is integrally closed, $sx \in A$. Hence $x \in A_S$. Hence $A_S$ is integrally closed.
Let $f$ be a non-zero element of $A_S$. $f = a/s$, where $a \in A, s \in S$. Then $fA_S = aA_S$. By this, $aA$ is a product of prime ideals of $A$. Let $P$ be a non-zero prime ideal $P$ of $A$. Since $P$ is maximal, $A_S/P^nA_S$ is isomorphic to $A/P^n$ or $0$. Hence $A_S/aA_S$ is a finite $k$-module. QED
Lemma 9 Let $A$ be an integrally closed domain containing a field $k$ as a subring. Suppose $A/fA$ is a finite $k$-module for every non-zero element $f \in A$. Let $P$ be a non-zero prime ideal of $A$. Then $A_P$ is a discrete valuation ring.
Proof: By Lemma 8 and this, every non-zero ideal of $A_P$ has a unique factorization as a product of prime ideals. Hence $PA_P \neq P^2A_p$. Let $x \in PA_P - P^2A_P$. Since $A_P$ is the only non-zero prime ideal of $A_P$, $xA = PA_P$. Since every non-zero ideal of $A_P$ can be written $P^nA_P$, $A_P$ is a principal ideal domain. Hence $A_P$ is a discrete valuation ring. QED
Theorem Let $k$ be a field. Let $K$ be a finitely generated extension field of $k$ of transcendence degree one. Let $A$ be a subring of $K$ containing $k$. Let $P$ be a prime ideal of $A$. Then there exists a valuation ring $R$ of $K$ dominating $A_P$.
Proof: We can assume that $A$ contains a transcendental element $x$ over $k$(otherwise the theorem would be trivial). We can also assume that $P \neq 0$.
Let $B$ be the integral closure of $A$ in $K$. By Lemma 7, $B/fB$ is a finite $k$-module for every non-zero element $f \in B$. Let $S = A - P$. Let $B_P$ and $A_P$ be the localizations of $B$ and $A$ with rspect to $S$ respectively. Let $y \in P$ be a non-zer element. By Lemma 8, $B_P/yB_P$ is a finite k-module. Since $yB_P \subset PB_P$ and $PB_P \neq B_P$, $yB_P \neq B_P$. Hence there exists a maximal ideal $Q$ of $B_P$ containing $y$. Since $B_P$ is integral over $A_P$ and $PA_P$ is a unique maximal ideal of $A_P$, $P = Q \cap A_P$. Let $Q' = Q \cap B$. Then $Q'$ is a prime ideal of $B$ lying over $P$. By Lemma 9, $B_Q'$ is a discrete valuation ring and it dominates $A_P$. QED