The other answer is correct, but a little brief so I decided to add a few words.
Let's begin with the Riesz representation: For a locally compact Hausdorff space $X$,
the map taking the finite signed regular Borel measure $\mu$ to the functional $f\mapsto \langle f,\mu\rangle:= \int f\,d\mu$
is an isometric isomorphism of the Banach space $M_r(X)$ onto the dual of the
Banach space $C_0(X)$. Here $C_0(X)$ is the space of continuous functions that vanish at infinity,
and the norm on $M_r(X)$ is the total variation norm.
Now Banach-Alaoglu guarantees that the unit ball in $C_0(X)^*$ is weak${}^*$ compact, which
corresponds to vague compactness for the measures. However, generally,
the unit sphere in $C_0(X)^*$ is not weak${}^*$ closed and hence not compact. Note however that the
cone $\cal P$ of positive elements in $M_r(X)$ is vaguely closed:
$${\cal P}=\bigcap_{f\in C_0(X)_+} \left\{\mu: \langle f,\mu\rangle \geq 0\right\},$$
so its intersection with the unit ball is compact.
The intuition here is that the positive part of the unit sphere is tiny, so
could well be compact even though the whole sphere itself is not.
So how does the compactness of $X$ enter into the picture? If $X$ is compact,
then $C_0(X)=C_b(X)$ and the vague and weak topologies of measures coincide.
In particular, the constant function "1" belongs to $C_0(X)$ so the space
of probability measures is the compact set
$$ {\cal P}\cap \{\mu: \|\mu\|\leq 1\}\cap \{\mu: \langle 1,\mu\rangle =1\}.$$
To put this result into context, I will show how to deduce it from the big theorems. (This is not nearly as elementary as Kavi's proof.)
Unfortunately, different authors use different notation / terminology, so let me set up a few definitions first.
Definition. A finitely additive signed measure (or charge) is a function $\mu : \mathcal A \to [-\infty,\infty]$ with the following properties:
- $\mu(\varnothing) = 0$;
- $\mu$ assumes at most one of the values $-\infty$ and $+\infty$;
- $\mu$ is finitely additive.
Comparted to signed measures, the requirement of $\sigma$-additivity is weakened to finite additivity.
If $\Omega$ is a topological space, let $\mathcal A_\Omega$ and $\mathcal B_\Omega$ denote respectively the algebra and the $\sigma$-algebra generated by the open (or closed) sets of $\Omega$. Then $\mathcal A_\Omega \subseteq \mathcal B_\Omega$, and $\mathcal B_\Omega$ is the $\sigma$-algebra generated by $\mathcal A_\Omega$, so every measure on $B_\Omega$ is uniquely determined by its values on $A_\Omega$.
Definition. Let $\Omega$ be a topological space. We say that a finitely additive signed measure $\mu$ on $\mathcal A_\Omega$ is regular if for every $A \in \mathcal A_\Omega$ one has
\begin{align*}
\mu(A) &= \sup\{\mu(F) \, : \, F \subseteq A\ \text{closed}\} \\[1ex]
&= \inf\,\{\mu(V) \: : \: V \supseteq A \ \text{open}\}.
\end{align*}
Note. Different authors use different notions of regularity. In particular, sometimes the closed sets $F \subseteq A$ are replaced by compact sets.
We will use the following well-known results:
Theorem. Let $\Omega$ be a normal Hausdorff space. Then $C_b(\Omega)'$ is isometrically isomorphic to the space $rba(\Omega)$ of all regular, finitely additive signed measures of bounded variation on $\mathcal A_\Omega$, equipped with the total variation norm.
See [DS58, Theorem IV.6.2 (p.262)] or [AB06, Theorem 14.10 (p.495)].
For general topological spaces, see this question on MathOverflow.
Lemma. Every finite signed Borel measure on a metric space is regular.
See [DS58, Exercise III.9.22 (p.170)] or [AB06, Theorem 12.5 (p.436)], among others.
(Side note: this is not true for the other notion of regularity, with compact sets instead of closed sets, as can be seen from this answer on MathOverflow.)
It follows that the space $\mathcal M(\Omega)$ of finite signed Borel measures is a subspace of $rba(\Omega) \cong C_b(\Omega)'$. To complete the proof, note that every normed space $X$ separates points on every subspace of $X'$: if $\varphi(x) = 0$ for all $x \in X$, then $\varphi = 0$.
(More generally, for a bilinear pairing $\langle E , F \rangle$ to be non-degenerate, so that it is a proper dual pairing, it is necessary and sufficient that the induced maps $E \to F^*$ and $F \to E^*$ are injective.)
References.
[DS58] Nelson Dunford, Jacob T. Schwartz, Linear Operators, Part I: General Theory, Interscience, 1958.
[AB06] Charalambos D. Aliprantis, Kim C. Border, Infinite Dimensional Analysis, A Hitchhiker's Guide, Third Edition, Springer, 2006.
Best Answer
Lemma. If $\mu(F) = \nu(F)$ for all closed subsets $F$ of $E$, then $\mu = \nu$.
Sketch of the proof. The set of all subsets $A$ of $E$ such that $\mu(A) = \nu(A)$ is a $\lambda$-system. It contains the closed subsets, which form a $\pi$-system that generates the Borel $\sigma$-algebra. Use the $\pi-\lambda$ theorem to conclude.
Lemma If $\int f \mathrm{d}\mu = \int f \mathrm{d}\nu$ for every $f \in C_b(E)$, then $\mu(F) = \nu(F)$ for all closed subsets $F$ of $E$.
Sketch of the proof. Let $F \subset E$ be closed. For each integer $n \geqslant 1$, let $f_n \colon E \to \mathbb{R}$ be the function defined by $f_n(x) = \min\{n d(x,F), 1\}$. Then $f_n$ is in $C_b(E)$ and the sequence $(f_n)$ converges pointwise to $\mathbf{1}_F$, and is decreasing: $f_n(x) \downarrow \mathbf{1}_F(x)$. By the Beppo-Levi theorem (monotone convergence), $\int f_n \mathrm{d}\mu \to \mu(F)$ and $\int f_n \mathrm{d}\nu \to \nu(F)$.