In my previous answer, we used a fact that an invertible ideal is projective and a fact that a finitely generated projective module over a local ring is free.
Here is a proof without using these facts.
Lemma 1
Let $A$ be a Noetherian local domain.
Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal.
Let $K$ be the field of fractions of $A$.
Let $\mathfrak{m}^{-1} = \{x \in K; x\mathfrak{m} ⊂ A\}$.
Then $\mathfrak{m}^{-1} \neq A$.
Proof:
Let $a \neq 0$ be an element of $\mathfrak{m}$.
By the assumption, Supp$(A/aA) = \{\mathfrak{m}\}$.
Since Ass$(A/aA) \subset$ Supp($A/aA)$, Ass$(A/aA) = \{\mathfrak{m}\}$.
Hence there exists $b \in A$ such that $b \in A - aA$ and $\mathfrak{m}b \subset aA$.
Since $\mathfrak{m}(b/a) \subset A$, $b/a \in \mathfrak{m}^{-1}$.
Since $b \in A - aA$, $b/a \in K - A$.
QED
Lemma 1.5
Let $A$ be an integral domain.
Let $K$ be the field of fractions of $A$.
Let $M \neq 0$ be a finitely generated $A$-submodule of $K$.
Let $x \in K$ be such that $xM \subset M$.
Then $x$ is integral over $A$.
Proof:
Let $\omega_1,\dots,\omega_n$ be generators of $M$ over $A$.
Let $x\omega_i = \sum_j a_{i,j} \omega_j$.
Then $x$ is a root of the characteristic polynomial of the matrix $(a_{ij})$.
QED
Lemma 2
Let $A$ be an integrally closed Noetherian local domain.
Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal.
Then $\mathfrak{m}$ is invertible.
Proof:
Let $K$ be the field of fractions of $A$.
Let $a \neq 0$ be an element of $\mathfrak{m}$.
Let $\mathfrak{m}^{-1} = \{x \in K; x\mathfrak{m} \subset A\}$.
Since $\mathfrak{m} \subset \mathfrak{m}\mathfrak{m}^{-1} \subset A$,
$\mathfrak{m}\mathfrak{m}^{-1} = \mathfrak{m}$ or $\mathfrak{m}\mathfrak{m}^{-1} = A$.
Suppose $\mathfrak{m}\mathfrak{m}^{-1} = \mathfrak{m}$.
Since $\mathfrak{m}$ is finitely generated, every element of $\mathfrak{m}^{-1}$ is integral over $A$ by Lemma 1.5.
Since $A$ is integrally closed, $\mathfrak{m}^{-1} \subset A$.
This is a contradiction by Lemma 1.
Hence $\mathfrak{m}\mathfrak{m}^{-1} = A$.
QED
Lemma 3
Let $A$ be a Noetherian local domain.
Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal.
Then $\bigcap_n \mathfrak{m}^n = 0$.
Proof:
Let $I = \bigcap_n \mathfrak{m}^n$.
Suppose $I \neq 0$.
Since dim$(A/I) = 0$, $A/I$ is an Artinian ring.
Hence there exists $n$ such that $\mathfrak{m}^n \subset I$.
Since $I \subset \mathfrak{m}^n$, $I = \mathfrak{m}^n$.
Since $I \subset \mathfrak{m}^{n+1}$, $\mathfrak{m}^n = \mathfrak{m}^{n+1}$.
By Nakayama's lemma, $\mathfrak{m}^n = 0$.
Hence $I = 0$.
This is a contradiction.
QED
Lemma 4
Let $A$ be an integrally closed Noetherian local.
Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal.
Let $I$ be a non-zero ideal of $A$ such that $I \neq A$.
Then $I = \mathfrak{m}^n$ for some integer $n > 0$.
Proof:
By Lemma 3, there exists $n > 0$ such that $I \subset \mathfrak{m}^n$ and I is not contained in $\mathfrak{m}^{n+1}$.
By Lemma 2, $\mathfrak{m}$ is invertible.
Since $I \subset \mathfrak{m}^n$, $I\mathfrak{m}^{-n} \subset A$.
Suppose $I\mathfrak{m}^{-n} \neq A$.
Then $I\mathfrak{m}^{-n} \subset \mathfrak{m}$.
Hence $I \subset \mathfrak{m}^{n+1}$.
This is a contradiction.
Hence $I\mathfrak{m}^{-n} = A$.
Hence $I = \mathfrak{m}^n$.
QED
Theorem
Let $A$ be an integrally closed Noetherian local.
Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal.
Then $A$ is a discrete valuation ring.
Proof:
By Nakayama's lemma, $\mathfrak{m} \neq \mathfrak{m}^2$.
Let $x \in \mathfrak{m} - \mathfrak{m}^2$.
By Lemma 4, $xA = \mathfrak{m}$.
Let $I$ be a non-zero ideal of $A$ such that $I \neq A$.
By Lemma 4, $I = \mathfrak{m}^n$.
Hence $I$ is principal.
Hence $A$ is a discrete valuation ring.
QED
Let $R=\cap_{\lambda\in\Lambda}R_{\lambda}$ with $R_{\lambda}$ DVRs (as in Matsumura's definition of Krull domains). Assume that the intersection is irredundant, that is, if $\Lambda'\subsetneq\Lambda$ then $\cap_{\lambda\in\Lambda}R_{\lambda}\subsetneq\cap_{\lambda'\in\Lambda}R_{\lambda'}$.
Let's prove that $m_{\lambda}\cap R$ is a prime ideal of height one, for all $\lambda\in\Lambda$. First note that $m_{\lambda}\cap R\neq (0)$. If the height of some $m_{\alpha}\cap R$ is at least $2$, then there exists a nonzero prime $p\subsetneq m_{\alpha}\cap R$. From Kaplansky, Commutative Rings, Theorem 110, there exists $m_{\alpha'}\cap R\subseteq p$ (obviously $\alpha'\neq\alpha$). Let $x\in\cap_{\lambda\ne\alpha}R_{\lambda}$, $x\notin R$ (so $x\notin R_{\alpha}$), and $y\in m_{\alpha'}\cap R$, $y\neq 0$. One can choose $m,n$ positive integers such that $z=x^my^n$ is a unit in $R_{\alpha}$. Since $x\in R_{\alpha'}$ and $y\in m_{\alpha'}\cap R$ we get $z\in m_{\alpha'}$. Obviously $z\in R_{\lambda}$ for all $\lambda\ne \alpha, \alpha'$, so $z\in R$, $z$ is invertible in $R_{\alpha}$ and not invertible in $R_{\alpha'}$, a contradiction with $m_{\alpha'}\cap R\subseteq m_{\alpha}\cap R$.
(This argument is adapted from Kaplansky's proof of Theorem 114. Furthermore, using again Theorem 110 one can see that $m_{\lambda}\cap R$ are the only height one prime ideals of $R$.)
Best Answer
The ideals of a DVR are all of the form $(\pi^n)$ where $\pi$ is any element with valuation $1$ (let's assume that the image of $v$ is $\mathbb{Z}$ and not $k\mathbb{Z}, k >1$).
To see this, suppose $I$ is an ideal and pick $a \in I$ with minimal valuation $n$. Then $v(a/\pi^n) = 0$ so $a/\pi^n$ is a unit, so $\pi^n \in I$. This means that $(\pi^n) \subseteq I$.
Conversely, if $b \in I, b \neq 0$ then $v(b) \ge v(a)$ so $v(b/\pi^n) \ge 0$. Therefore $b/\pi^n \in R$ and therefore $b = (b/\pi^n) \cdot \pi^n \in (\pi^n)$.
So given that the ideals of $A$ are all of the form $(\pi^n)$, we immediately get unique factorization of (fractional) ideals because $(\pi^n) = (\pi)^n$.
The ideal $(\pi)$ is maximal since $x \in (\pi) \iff v(x) \ge 1$ so $A - (\pi) = \{x : v(x) = 0\} = A^\times$.