The underlying metric space is separable if and only if the space of finite Borel measures is separable in Prokhorov metric I'm trying to prove below result about Prokhorov metric. Could you verify if my attempt is fine?

Let $(X, d)$ be a metric space and $\mathcal{M} :=\mathcal{M}(X)$ the set all non-negative finite Borel measures on $X$. Let $d_P$ be the Prokhorov metric on $\mathcal{M}$.

Theorem: $(X, d)$ is separable if and only if $(\mathcal{M}, d_P)$ is.


I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.
 A: *

*Lemma 1: If $X$ is separable, then convergence in $d_P$ is equivalent to weak convergence.


*Lemma 2: Let $\mu, \mu_1,\mu_2,\ldots \in \mathcal M$. Then $\mu_i \to \mu$ weakly if and only if $\int f \mathrm d \mu_i \to \int f \mathrm d \mu$ for all uniformly continuous and bounded functionals $f$.
Notice that $x \mapsto \delta_{x}$ is a homeomorphism from $X$ onto $\left\{\delta_{x} \mid x \in X\right\}$. If $\mathcal{M}$ is separable, then so is $\left\{\delta_{x} \mid x \in X\right\}$ and thus is $X$. Let's prove the other direction. Let $D$ be a countable dense subset of $X$. Let
$$
\mathcal D := \{\alpha_1 \delta_{a_1} + \cdots + \alpha_k \delta_k \mid a_1, \ldots, a_k \in D \text{ and }\alpha_1,\ldots, \alpha_k \in \mathbb Q_{\ge 0}\}
$$
Then $\mathcal D$ is countable. Let's prove that $\mathcal D$ is dense in $\mathcal{M}$. Fix $\mu \in \mathcal{M}$. For each $m\ge 1$, we pick $k_m$ such that
$$
\mu \left ( \bigcup_{j=1}^{k_m} B(a_j, 1/m) \right )  \ge \mu(X)-1/m.
$$
Let $A_{1}^{m} := B\left(a_{1}, 1 \right)$ and
$$
A_{j}^{m} := B\left(a_{j}, 1 / m\right) \setminus \bigcup_{i=1}^{j-1} B\left(a_{i}, 1 / m\right) \quad \forall j=2, \ldots, k_{m}.
$$
Then $(A_j^m)_{j=1}^{k_m}$ is disjoint, and their union is equal to $\bigcup_{i=1}^{k_m} B\left(a_{i}, 1 / m\right)$ for all $m\ge 1$. In particular,
$$
\mu(X) \ge\sum_{j=1}^{k_{m}} \mu\left(A_{j}^{m}\right) \ge \mu(X)-1/m \quad \forall m \ge 1.
$$
We approximate
$$
\mu\left(A_{1}^{m}\right) \delta_{a_{1}}+\cdots+\mu\left(A_{k_{m}}^{m}\right) \delta_{a_{k_{m}}} \quad \text{by} \quad
\mu_{m}:=\alpha_{1}^{m} \delta_{a_{1}}+\cdots+\alpha_{k_{m}}^{m} \delta_{a_{k_{m}}}
$$
such that $\alpha_{1}^{m}, \ldots, \alpha_{k_{m}}^{m} \in \mathbb Q_{\ge 0}$ and
$$
\sum_{j=1}^{k_{m}} \left|\mu\left(A_{j}^{m}\right)-\alpha_{j}^{m}\right|<2 / m.
$$
Let $g$ be a uniformly continuous and bounded functional on $X$. By Lemmas 1 and 2, we need to prove $\int g \mathrm d \mu_m \to \int g \mathrm d \mu$ as $m \to \infty$. In deed,
$$
\begin{align}
& \left |\int g \mathrm d \mu_m - \int g \mathrm d \mu \right | \\
= &\left | \sum_{j=1}^{k_{m}}\alpha_j^m g(a_j) - \int g \mathrm d \mu \right| \\
\le & \left | \sum_{j=1}^{k_{m}} \mu\left(A_{j}^{m}\right) g(a_j) - \int g \mathrm d \mu \right| + \frac{2}{m} \sup_j | g(a_j) | \\
\le& \left | \int \sum_{j=1}^{k_{m}}  g(a_j) 1_{A_j^m} \mathrm d \mu - \int g \mathrm d \mu \right| + \frac{2}{m} \|g\|_\infty \\
=& \left | \int \sum_{j=1}^{k_{m}}  [g(a_j) -g] 1_{A_j^m} \mathrm d \mu  + \int g 1_{(\bigcup_{j=1}^{k_{m}} A_j^m)^c} \right| + \frac{2}{m} \|g\|_\infty\\
\le&  \sum_{j=1}^{k_{m}}  \int  | g(a_j) -g| 1_{A_j^m} \mathrm d \mu  +  \|g\|_\infty \mu \left ( \left (\bigcup_{j=1}^{k_{m}} A_j^m \right )^c \right ) + \frac{2}{m} \|g\|_\infty\\
\le&  \sum_{j=1}^{k_{m}}  \sup_{x\in A_j^m} | g(a_j) -g(x)| \mu (A_j^m) +  \frac{1}{m} \|g\|_\infty + \frac{2}{m} \|g\|_\infty.
\end{align}
$$
Each $A_{j}^{m}$ is contained in a ball with radius $1 / m$ around $a_{j}$. Since $g$ is uniformly continuous, for every $\varepsilon>0$ there is a $\delta>0$ such that $|g(y)-g(x)|<\varepsilon$ whenever $d(x,y)<\delta$, so $\left|g(a_{j}) - g(x)\right|<\varepsilon$ for all $j$ and $x \in A_{j}^{m}$. Then for $m$ such that $1/m < \min\{\varepsilon, \delta\}$, it follows from the above computation that
$$
\left|\int g \mathrm d \mu_{m}-\int g \mathrm  d \mu\right| \leq \varepsilon \mu(X) + \frac{3}{m} \|g\|_{\infty}.
$$
This completes the proof.
