Lemma 1: Let $\mu, \mu_1, \mu_2,\ldots \in \mathcal{M}$ and $g \in \mathcal C_b(X)$. If $\mu_i \to \mu$ weakly, then $\mu_i(A) \to \mu(A)$ for all Borel set $A \subseteq X$ with $\mu(\partial A) = 0$.
Lemma 2: If $X$ is separable and $\mu \in \mathcal{M}$. Then for each $\delta>0$ there are countably many open (or closed) balls $B_{1}, B_{2}, \ldots$ such that $\bigcup_{i=1}^{\infty} B_{i}=X$, the radius of $B_{i}$ is less than $\delta$, and $\mu\left(\partial B_{i}\right)=0$ for all $i$.
Fix $\varepsilon>0$. We want to show that $\exists N, \forall i \geq N: d_{P}\left(\mu_{i}, \mu\right) \leq \varepsilon$, i.e., $\mu_{i}(B) \leq \mu\left(B_{\varepsilon}\right)+\varepsilon$ and $\mu(B) \leq \mu_{i}\left(B_{\varepsilon}\right)+\varepsilon$ for all Borel subset $B$.
Fix $\delta \in (0, \varepsilon/4)$. By Lemma 2, there are countably many open balls $B_{1}, B_{2}, \ldots$ with radius less than $\delta/2$ such that $\bigcup_{i=1}^{\infty} B_{i}=X$ and $\mu\left(\partial B_{i}\right)=0$ for all $i$. Fix $k$ such that
$$
\mu\left(\bigcup_{j=1}^{k} B_{j}\right) \ge \mu(X)-\delta.
$$
Let $\mathcal A$ be the finite collection of subsets built by combining the balls $B_1, \ldots, B_k$, i.e.,
$$
\mathcal{A}:=\left\{\bigcup_{j \in I} B_{j} \,\middle\vert\, J \subset \{1, \ldots, k\}\right\}.
$$
We will use this collection to approximate any Borel set. For each $A \in \mathcal{A}, \partial A \subset \partial B_{1} \cup \cdots \cup \partial B_{k}$, so $\mu(\partial A) \leq$ $\mu\left(\partial B_{1}\right)+\cdots+\mu\left(\partial B_{k}\right)=0$. By Lemma 1, $\mu_{i}(A) \rightarrow \mu(A)$ for all $A \in \mathcal{A}$. Fix $N$ such that
$$
\left|\mu_{i}(A)-\mu(A)\right|<\delta \quad \forall i \geq N, \forall A \in \mathcal{A}.
$$
In particular,
$$
\mu_i \left(\bigcup_{j=1}^{k} B_{j}\right) \ge \mu \left(\bigcup_{j=1}^{k} B_{j}\right) -\delta \ge \mu(X) - 2 \delta \quad \forall i \ge N.
$$
Now we fix a Borel set $B$ and approximate it by
$$
A := \bigcup \{B_j \mid j = 1,\ldots,k \text{ such that } B_j \cap B \neq \emptyset\}.
$$
Then
- $A \subset B_{\delta} := \{x \mid d(x, B)<\delta\}$ because $\operatorname{diam} B_{j}<\delta$,
- $B=\left[B \cap \bigcup_{j=1}^{k} B_{j}\right] \cup\left[B \cap\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right] \subset \left [ A \cup\left(\bigcup_{j=1}^{k} B_{j}\right)^{c} \right ]$,
- $\left|\mu_{i}(A)-\mu(A)\right|<\delta$ for all $i \geq N$, and
- $\mu\left(\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right) \leq \delta$ and $\mu_{i}\left(\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right) \leq \mu_i(X)-\mu(X)+ 2 \delta \le \mu_i(X) + 3\delta$ for all $i \geq N$.
It follows that for every $i \geq N$ :
\begin{aligned}
\mu(B) & \leq \mu(A)+\mu\left(\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right) \\
& \leq \mu(A)+\delta \\
& \leq \mu_{i}(A)+2 \delta \\
& \leq \mu_{i}\left(B_{\delta}\right)+2 \delta \\
&\leq \mu_{i}\left(B_{\varepsilon}\right)+\varepsilon \\
\mu_{i}(B) & \leq \mu_{i}(A)+\mu_{i}\left(\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right) \\
&\leq \mu_{i}(A)+ 3 \delta \\
&\leq \mu(A)+4 \delta \\
& \leq \mu\left(B_{\delta}\right)+4 \delta \\
&\leq \mu\left(B_{\varepsilon}\right)+\varepsilon.
\end{aligned}
This is true for every $B \in \mathcal{B}$, so $d_{P}\left(\mu_{i}, \mu\right) \leq \varepsilon$ for all $i \geq N$.
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.
Best Answer
Lemma 1: Assume $D$ is countable and dense in $X$. Then $$ \mathcal D := \{a_1 \delta_{x_1} + \cdots + a_n \delta_{x_n} \mid n \in \mathbb N_{>0} \text{ and } x_1, \ldots, x_n \in D \text{ and } a_1, \ldots, a_n \in \mathbb Q_{\ge 0} \text{ and } a_1 + \cdots + a_n=1\} $$ is countable and dense in $\mathcal P$.
Lemma 2: Suppose $X$ has a countable dense subset $A$ such that every Cauchy sequence in $A$ converges to a point in $X$. Then $X$ is complete.
Lemma 3: Let $X$ be complete and $\Gamma \subset \mathcal{P}$. If $$ \forall \varepsilon, \delta>0, \exists a_{1}, \ldots, a_{n} \in X, \forall \mu \in \Gamma: \mu\left(\bigcup_{i=1}^{n} B\left(a_{i}, \delta\right)\right) \geq 1-\varepsilon, $$ then $\Gamma$ is uniformly tight.
Let $D$ be a countable dense subset of $X$. By Lemma 1, let $\mathcal D$ be a countable dense subset of $\mathcal P$ that is induced by $D$. Let $(\mu_n)$ with $\mu_n := a_{n,1} \delta_{x_{n,1}} + \cdots + a_{n, \lambda(n)} \delta_{x_{n, \lambda(n)}}$ be a Cauchy sequence in $\mathcal D$. By Lemma 2, it suffices to show that there is $\mu \in \mathcal P$ such that $\mu_n \to \mu$ in $d_P$. By Prokhorov theorem, it suffices to show that $(\mu_n)$ is uniformly tight. As such, we will prove that $(\mu_n)$ satisfies the condition of Lemma 3.
Fix $\varepsilon, \delta>0$. Let's $\varepsilon' := \frac{1}{2} \min\{\varepsilon, \delta\}$. There is $N$ such that $d_P(\mu_n, \mu_N) < \varepsilon'$ for all $n \ge N$, i.e., $$ \begin{cases} \mu_N(A) \leq \mu_n \left(A_{\varepsilon'}\right)+\varepsilon' \\ \mu_n(A) \leq \mu_N \left(A_{\varepsilon'}\right)+\varepsilon' \end{cases} \quad \forall n \ge N, A \in \mathcal B(X). $$
Let $$ A := \bigcup_{i=1}^{\lambda(N)} B (a_{N,i}, \varepsilon' ). $$
It follows that $\mu_n \left(A_{\varepsilon'}\right) \ge 1-\varepsilon'$ for all $n \ge N$. Notice that $$ A_{\varepsilon'} = \{x\in X \mid d(x, A) < \varepsilon'\} \subset \bigcup_{i=1}^{\lambda(N)} B (a_{N,i}, 2\varepsilon' ) \subset \bigcup_{i=1}^{\lambda(N)} B (a_{N,i}, \delta). $$
It follows that $$ \mu_n \left( \bigcup_{i=1}^{\lambda(N)} B (a_{N,i}, \delta) \right ) \ge \mu_n(A_{\varepsilon'}) \ge 1- \varepsilon' \ge 1-\varepsilon \quad \forall n \ge N. $$
It follows that the collection of centers $\{a_{n,i} \mid n \le N, i\le \lambda(n)\}$ satisfies the condition.