Let $\mathcal{F}$ be a rank $1$ coherent reflexive $\mathcal{O}_X$-module. Choose an open $U \subset X$ such that every irreducible component of $X \setminus U$ has codimension $\geq 2$ in $X$ and such that $\mathcal{F}|_U$ is invertible, see Lemma \ref{lemma-reflexive-over-normal}. Observe that $\text{Cl}(U) = \text{Cl}(X)$ as the Weil divisor class group of $X$ only depends on its field of rational functions and the points of codimension $1$ and their local rings. Thus we can define the Weil divisor class of $\mathcal{F}$ to be the Weil divisor class of $\mathcal{F}|_U$ in $\text{Cl}(U)$. We omit the verification that this is independent of the choice of $U$. \medskip\noindent Denote $\text{Cl}'(X)$ the set of isomorphism classes of rank $1$ coherent reflexive $\mathcal{O}_X$-modules. The construction above gives a group homorphism $$ \text{Cl}'(X) \longrightarrow \text{Cl}(X) $$ because for any pair $\mathcal{F}, \mathcal{G}$ of elements of $\text{Cl}'(X)$ we can choose a $U$ which works for both and the assignment (\ref{equation-c1}) sending an invertible module to its Weil divisor class is a homorphism. If $\mathcal{F}$ is in the kernel of this map, then we find that $\mathcal{F}|_U$ is trivial (Lemma \ref{lemma-normal-c1-injective}) and hence $\mathcal{F}$ is trivial too by Lemma \ref{lemma-reflexive-S2-extend} and Serre's criterion Properties, Lemma \ref{properties-lemma-criterion-normal}. To finish the proof it suffices to check the map is surjective. \medskip\noindent Let $D = \sum n_Z Z$ be a Weil divisor on $X$. We claim that there is an open $U \subset X$ such that every irreducible component of $X \setminus U$ has codimension $\geq 2$ in $X$ and such that $Z|_U$ is an effective Cartier divisor for $n_Z \not = 0$. To prove the claim we may assume $X$ is affine. Then we may assume $D = n_1 Z_1 + \ldots + n_r Z_r$ is a finite sum with $Z_1, \ldots, Z_r$ pairwise distinct. After throwing out $Z_i \cap Z_j$ for $i \not = j$ we may assume $Z_1, \ldots, Z_r$ are pairwise disjoint. This reduces us to the case of a single prime divisor $Z$ on $X$. As $X$ is $(R_1)$ by Properties, Lemma \ref{properties-lemma-criterion-normal} the local ring $\mathcal{O}_{X, \xi}$ at the generic point $\xi$ of $Z$ is a discrete valuation ring. Let $f \in \mathcal{O}_{X, \xi}$ be a uniformizer. Let $V \subset X$ be an open neighbourhood of $\xi$ such that $f$ is the image of an element $f \in \mathcal{O}_X(V)$. After shrinking $V$ we may assume that $Z \cap V = V(f)$ scheme theoretically, since this is true in the local ring at $\xi$. In this case taking $$ U = X \setminus (Z \setminus V) = (X \setminus Z) \cup V $$ gives the desired open, thereby proving the claim. \medskip\noindent In order to show that the divisor class of $D$ is in the image, we may write $D = \sum_{n_Z < 0} n_Z Z - \sum_{n_Z > 0} (-n_Z) Z$. By additivity of the map constructed above, we may and do assume $n_Z \leq 0$ for all prime divisors $Z$ (this step may be avoided if the reader so desires). Let $U \subset X$ be as in the claim above. If $U$ is quasi-compact, then we write $D|_U = -n_1 Z_1 - \ldots - n_r Z_r$ for pairwise distinct prime divisors $Z_i$ and $n_i > 0$ and we consider the invertible $\mathcal{O}_U$-module $$ \mathcal{L} = \mathcal{I}_1^{n_1} \ldots \mathcal{I}_r^{n_r} \subset \mathcal{O}_U $$ where $\mathcal{I}_i$ is the ideal sheaf of $Z_i$. This is invertible by our choice of $U$ and Lemma \ref{lemma-sum-effective-Cartier-divisors}. Also $\text{div}_\mathcal{L}(1) = D|_U$. Since $\mathcal{L} = \mathcal{F}|_U$ for some rank $1$ coherent reflexive $\mathcal{O}_X$-module $\mathcal{F}$ by Lemma \ref{lemma-reflexive-S2-extend} we find that $D$ is in the image of our map. \medskip\noindent If $U$ is not quasi-compact, then we define $\mathcal{L} \subset \mathcal{O}_U$ locally by the displayed formula above. The reader shows that the construction glues and finishes the proof exactly as before. Details omitted.