Let $M$ be a metric space. Then we define the space $\mathcal M_1(M)$ as the topological space
$$\mathcal M_1(M) :=\left\{\mu;\ \mu\ \text{is a } \text{Borel probability measure on }M\right\} $$ endowed with the weak$^*$ topology, $\textit{i.e}.$ the topology generated by the basis of neighborhoods $$V(\mu;f_1,\ldots,f_n;\varepsilon):=\left\{\lambda\in\mathcal M_1(M);\ \left|\int f_i\ \mathrm{d}\mu - \int f_i\ \mathrm{d}\lambda\right|<\varepsilon, \ \forall \ i\in\{1,\ldots,n\}\right\}, $$ where $f_1,\ldots,f_n \in C^0_b(M)=\{g: M\to\mathbb R; \ g \text{ is a continuous bounded function}\}$.
It is well known that if $M$ is a separable metric space. Then $M$ is compact $\iff$ $\mathcal{M}_1(M)$ is compact. The proof of this fact relies on the result "$M$ is separable metric space then $\mathcal{M}_1(M)$ in the weak$^*$ topology is metrizable with Lévy–Prokhorov metric".
Question: If $M$ is a non-compact metric space $\Rightarrow$ $\mathcal M_1(M)$ a non-compact topological space in the weak$^*$ topology?
My conclusions so far
I think I have found a solution. Suppose that $M$ is a non-compact metric space then there exists a sequence $\{x_n\}_{n\in\mathbb{N}}$, such that, $\{x_n\}_{n\in\mathbb{N}}$ does not admit convergent subsequênce. Therefore, the set $S = \{x_1, \ldots, x_i,\ldots \}\subset M$ is closed, moreover all subsets of $S$ are closed as well.
Now, consider the sequence $\{\delta_{x_i}\}_{i\in\mathbb{N}}\subset\mathcal M_1(M)$ (where $\delta_p$ is the Dirac measure), we will show that $\{\delta_{x_i}\}_{i\in\mathbb{N}}$ does not admit convergence subnet, and this will imply that $\mathcal M_1(M)$ is not compact in the weak$^*$ topology (because we would be able to find a net that does not admit convergent subnet).
Suppose by reductio ad absurdum that there exist a directed set $(\mathcal B, \preccurlyeq)$, an increasing function $\varphi: \mathcal B\to\mathbb{N}$, such that $\lim_{\beta}\varphi(\beta)=\infty$ (or cofinal porperty) and $\mu\in\mathcal M_1(M)$ satisfying $$\lim_{\beta} \delta_{x_{\varphi(\beta)}} =\mu \in\mathcal M_1(M). $$
Let us, first of all, prove that $\mu(S)=1$. Let $f: M\to\mathbb{R}$ a continuous bounded function such that $f(S)=\{1\}.$ Then, by the definition of weak$^*$ topology $$1 = \lim_\beta \int f \ \mathrm d \delta_{x_{\varphi(\beta)}} = \int f \ \mathrm d \mu.$$
Now consider the continuous (since $S$ is closed) bounded functions \begin{align*} g_n: M&\to\mathbb{[0,1]}\\ x&\mapsto \max\left\{ 1 - n \cdot d(x,S) ,0\right\}, \end{align*} so $g_n \to \mathbf{1}_S$ pointwiselly ( where $\mathbf{1}_S( S ) =\{1\}$ and $\mathbf{1}_S\left(M\setminus S\right) =\{0\}$) and each $g_n$ is a continuous bounded function such that $g_n(S)=\{1\}$. Note that $|g_i(x)|\leq g_1(x)$, $\forall$ $x\in M$. Thus, by dominated convergence theorem $$1 = \lim_{n\to\infty} \int g_n\ \mathrm{d}\mu = \int \lim_{n\to\infty}g_n\ \mathrm{d}\mu = \int \mathbf{1}_s\ \mathrm{d}\mu = \mu(S). $$
Now, let $i\in\mathbb{\mathbb N}$ and consider $S_i = \{1,\ldots,i\}$, by hypothesis $S_i$ and $S\setminus S_i$ are both closed, so there exists a continuous function $g: M\to [0,1] $ such that $g(S_i) = \{1\}$ and $g(S \setminus S_i) = \{0\}$, since $\varphi(\beta)\to\infty$, there exists $\beta_0\in\mathcal B$, $\beta_0\preccurlyeq \beta$, implies $i < \varphi(\beta)$. Hence, $$0 = \lim_\beta \int g \ \mathrm d \delta_{x_{\varphi(\beta)}} = \int g \ \mathrm d \mu \geq \mu(S_i).$$
Since $\mu(S) \leq \sum_{i=1}^{\infty}\mu(S_i) = 0$. Then $\mu\not\in \mathcal{M}_1(M)$, implying that $M$ is not compact.
Is the proof above correct? If so, why the vast majority of the books (at least all that I have checked) people just state the result for the case where $M$ is separable metric space?
Can anyone help me?