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.
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.