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