I assume that you have elementary knowledge of commutative algebra;
integral dependence, the lying-over theorem, localization of rings by multiplicative subsets, etc.(for example, see Atiyah-Macdonald).
I will prove a slightly more general theorem than yours.
Since a UFD is integrally closed, your theorem will follow immediately.
Theorem. Let $A$ be an integrally closed domain and let $P$ be a prime ideal of $A$. Let $K$ be the field of fractions of $A$. Let $\tilde{K}$ be the field of fractions of $A/P$. Let $f(X) \in A[X]$ be a monic polynomial without multiple roots. Let $\tilde{f}(X) \in (A/P)[X]$ be the reduction of $f(X)$ mod $P$. Suppose $\tilde{f}(X)$ is also without multiple roots. Let $L$ be the splitting field of $f(X)$ over $K$. Let $G$ be the Galois group of $L/K$. Let $\tilde{L}$ be the splitting field of $\tilde{f}(X)$ over $\tilde{K}$. Let $\tilde{G}$ be the Galois group of $\tilde{L}/\tilde{K}$. Then $\tilde{G}$ is isomorphic to a subgroup of $G$.
To prove this theorem, we need some notations.
Definition. Let $A$ be a ring. Let $G$ be a group. Let $\mathrm{Aut}(A)$ be the automorphism group of $A$. Suppose there exists a homomorphism $\psi：G \rightarrow \mathrm{Aut}(A)$. We say $G$ acts on $A$. For $\sigma \in G$ and $x \in A$, we denote $\psi(\sigma)(x)$ by $\sigma x$.
We denote the set $\{x \in A；\sigma x = x$ for all $\sigma \in G\}$ by $A^G$. Then $A^G$ is a subring of $A$.
The proof of the theorem.
Let $\alpha_1, ..., \alpha_n$ be the roots of $f(X)$ in $L$.
Let $B = A[\alpha_1, ..., \alpha_n]$.
Since $f(X)$ is monic, $B$ is integral over A.
Let $S = A \setminus P$.
Let $A_S$ and $B_S$ be localizations of $A$ and $B$ respectively.
Then $B_S = A_S[\alpha_1, ..., \alpha_n]$.
Since $A$ is integrally closed, $A_S$ is also integrally closed.
By replacing $A$ and $P$ by $A_S$ and $PA_S$ respectively, we can assume $P$ is a maximal ideal of $A$.
By the lying-over theorem, there exists a prime ideal $M$ of $B$ such that $P = A \cap M$.
Since $B$ is integral over $A$ and $P$ is a maximal ideal of $A$, $M$ is a maximal ideal of $B$.
Obviously $G$ acts on $B$. $A \subset B^G \subset K$ and $B^G$ is integral over $A$. Since $A$ is integrally closed, $B^G = A$.
Let $\pi:B → B/M$ be the canonical map. Then $B/M = (A/P)[\pi(\alpha_1), ..., \pi(\alpha_n)] = \tilde{K}(\pi(\alpha_1), ..., \pi(\alpha_n))$. Since $\pi(\alpha_1), ..., \pi(\alpha_n)$ are all the roots of $\tilde{f}(X)$, $B/M$ can be identified with $\tilde{L}$. Since $\tilde{f}(X)$ is without multiple roots, $B/M$ is separable over $A/P$. Hence $B/M$ has a primitive element $\theta$ over $A/P$. Let $y$ be an element of $B$ such that $\pi(y) = \theta$. Let $H = \{\sigma \in G; \sigma(M) = M\}$. Each $\sigma \in H$ induces $\tilde{\sigma} \in \tilde{G}$. Hence we get a homomorphism $\psi: H \rightarrow \tilde{G}$.
It suffices to prove that $\psi$ is an isomorphism. For each $\sigma \in G - H, \sigma(M) \neq M$. Hence, by the Chinese remainder theorem, there exists $x \in B$ such that $x \equiv y\bmod M$, $x \equiv 0\bmod\sigma(M)$ for each $\sigma \in G - H$. Let $F(X) = \prod_{\sigma \in G}(X - \sigma(x)) \in B[X]$. Since each coefficient of $F(X)$ is invariant by $G, F(X) \in (B^G)[X] = A[X]$. Let $\tilde{F}(X) = \prod_{\sigma \in G}(X - \pi(\sigma(x))$. Then $\tilde{F}(X) \in (A/P)[X]$. Let $\lambda$ be any element of $\tilde{G}$. Since $\pi(x)$ is a root of $\tilde{F}(X)$, $\lambda(\pi(x))$ is also a root of $\tilde{F}(X)$.
Hence there exists $\tau \in G$ such that $\lambda(\pi(x)) = \pi(\tau(x))$.
For each $\sigma \in G - H$, $\sigma(x) \in M$.
Hence $\pi(\sigma(x)) = 0$.
On the other hand, $\pi(\tau(x)) = \tilde{\tau}(\pi(x)) = \tilde{\tau}(\theta) \neq 0$
Hence $\tau \in H$.
Since $\lambda(\theta) = \pi(\tau(x)) = \tilde{\tau}(\pi(x)) = \tilde{\tau}(\theta)$,
$\lambda = \tilde{\tau}$.
Therefore $\psi: H \rightarrow \tilde{G}$ is surjective.
It remains to show that $\psi$ is injective.
Suppose $\psi$ is not injective.
Then there exists $\sigma \in H$ such that $\sigma \neq 1$ and $\psi(σ) = 1$.
Hence there exists a root $\alpha$ of $f(X)$ such that $\sigma(\alpha) \neq \alpha$. Since $\sigma(\alpha)\equiv\alpha\bmod M$, $\tilde{f}(X)$ has a multiple root $\pi(\alpha)$. This is a contradiction. QED
