Lemma: Let $f:X \to Y$ and $C$ be some collection of subsets of $Y$. Then $$f^{-1}[\sigma(C)]=\sigma(f^{-1}[C]).$$
Let $\mathcal O$ be the standard topology of $\mathbb R$. Then $\sigma(\mathcal O)$ is the Borel $\sigma$-algebra of $\mathbb R$. Let $\mathcal C(E)$ the space of all all continuous functionals on $E$. Let
$$
f^{-1}[\sigma(\mathcal O)] := \{f^{-1}(B) \mid B \in \sigma(\mathcal O)\} \quad \forall f \in \mathcal C(E).
$$
Then $\mathcal A$ is the $\sigma$-algebra generated by
\begin{align}
\bigcup_{f \in \mathcal C(E)} f^{-1}[\sigma(\mathcal O)].
\end{align}
We have
\begin{align}
\mathcal A &= \sigma \left ( \bigcup_{f \in \mathcal C(E)} f^{-1}[\sigma( \mathcal O )] \right ) \\
&= \sigma \left ( \bigcup_{f \in \mathcal C(E)} \sigma( f^{-1}[\mathcal O] ) \right ) \quad \text{by our Lemma} \\
& \subseteq \sigma \left ( \sigma \left ( \bigcup_{f \in \mathcal C(E)} f^{-1}[\mathcal O] \right) \right ) \\
&= \sigma \left ( \bigcup_{f \in \mathcal C(E)} f^{-1}[\mathcal O] \right) \\
&\subseteq \mathcal B(E).
\end{align}
Because metric space is perfectly normal. For each closed subset $F$ of $E$, there is a continuous functional $f:E \to [0, 1]$ such that $F =f^{-1}(0)$. This implies $\mathcal A$ contains all closed and thus all open subsets of $E$. This completes the proof.
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 $i \geq N$.
Best Answer
Let $D := (a_i)_{i=1}^\infty$ be a countable dense subset of $X$. For each $m \ge 1$, $$ \bigcup_{i=1}^\infty B \left(a_{i}, 1/m\right) = X. $$ Fix $\varepsilon > 0$. For each $m \ge 1$, there is a natural number $k_m$ such that $$ \mu \left ( \bigg (\bigcup_{i=1}^{k_m} B \left(a_{i}, 1/m\right) \bigg )^c \right ) \le2^{-m} \varepsilon. $$
Let $$ K := \bigcap_{m=1}^\infty \bigcup_{i=1}^{k_m} \overline B \left(a_{i}, 1/m\right). $$
Clearly, $K$ is closed and thus complete. Moreover, $K$ is totally bounded, so it's compact. We have
$$ \mu(K^c) \le \varepsilon \sum_{m=1}^\infty 2^{-m} = \varepsilon. $$
This completes the proof.