I'm trying to prove this well-known property. Could you verify if my attempt is fine?
Let $(X, d)$ be a metric space, $\operatorname{Lip}_{b}(X)$ the space of Lipschitz continuous bounded real-valued maps on $X$, and $\mu$ a finite Borel measure on $X$. Then $\operatorname{Lip}_{b}(X)$ is dense in $L_1(X, \mu)$ w.r.t. $\|\cdot\|_{L_1}$.
I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.
Let $f$ be a simple function, i.e., $f = \sum_{k=1}^n r_k 1_{B_k}$ where $r_k \in \mathbb R_{\neq 0}$ and $(A_k)_{k=1}^n$ is a finite sequence of pairwise disjoint sets in $\mathcal B(X)$. For each $k$, there is $f_k \in \operatorname{Lip}_{b}(X)$ such that $\|f_k-1_{B_k}\|_{L_1} < \varepsilon /(k|r_k|)$. This implies $\|r_kf_k-r_k1_{B_k}\|_{L_1} < \varepsilon / k$. Hence $$ \left \| f- \sum_{k=1}^n r_kf_k \right \|_{L_1} \le \varepsilon. $$
It follows that $\operatorname{Lip}_{b}(X)$ is dense in the space $S(X,\mu)$ of simple functions. On the other hand, $S(X,\mu)$ is dense in $L_1(X, \mu)$. The claim then follows.
We have a direct consequence, i.e.,
Let $(X, d)$ be a metric space and $\mu, \nu$ finite Borel measures on $X$. If $$ \int f \mathrm d \mu=\int f \mathrm d \nu \quad \text {for all} \quad f \in \operatorname{Lip}_{b}(X), $$ then $\mu=\nu$.