Here's one way to do it without getting your hands too dirty. It's not constructive.
Let ${\cal A}$ be the union of a countable dense set in $C_0({\mathbb R}^n)$, and indicators of all balls of integer radius centered at the origin (not necessary, but helpful later).
Let $\{X_1,X_2,\dots \}$ be an IID $\mu$-distributed sequence.
Let $\mu_n = \frac{1}{n} \sum_{j\le n} \delta_{X_j}.$
Then $\mu_n$ is a random probability measure. Observe that for any bounded and measurable function,
$$\int f d\mu_n= \frac {1}{n} \sum_{j\le n } f(X_j).$$
Then by LLN
$$ \lim_{n\to\infty} \int f d\mu_n = \int f d\mu ,~\forall f \in {\cal A}\quad (*),$$
a.s. (here is where we use the countability of ${\cal A}$).
Let $\{\bar \mu_n\}$ be a realization of $\{\mu_n\}$ (or equivalently, $\{X_1,\dots\}$) for which $(*)$ holds. It remains to show that $\{\bar \mu_n\}$ converges weakly.
Let $f\in C_b({\mathbb R}^d)$, by a standard procedure, for every $M$, there exists $f_M\in C_0({\mathbb R}^d)$ such that $f_M$ coincides with $f$ on $\{x:|x|<M\}$ and $|f_M |\le |f|$. Fix $\epsilon>0$ and choose an integer $M$ such that $\mu(\{x:|x|>M\})<\epsilon$
There exists a continuous $g_M\in {\cal A}$ such that $\|f_M-g_M\|<\epsilon$.
We then have
$$ \int f d \bar \mu_n = \int g_M d \bar \mu_n + \int (f_M -g_M) d \bar \mu_n +
\int (f- f_M) d \bar \mu_n.$$
The first integral on the RHS converges to $\int g_M d\mu$ by construction. The second is bounded in absolute value by $\epsilon$. The third is bounded by $2\|f\|\mu_n (\{x:|x|>M\})$, which by assumption (recall the indicator functions in ${\cal A}$) converges to $2\|f\|\mu(\{x:|x|>M\})=2\|f\|\epsilon$.
Therefore
$$ \limsup_{n\to\infty} |\int f d\bar \mu_n - \int g_M d\mu| \le \epsilon (1+2\|f\|).$$
But
$$\begin{align*} |\int g_M d\mu - \int f d \mu| &\le \int |g_M -f_M| d \mu + \int |f_M -f| d\mu\\
& \le \epsilon + 2\|f\|\mu (\{x:|x|>M\})\le \epsilon (1+2\|f\|)\end{align*}.$$
The result follows.
$\newcommand{\A}{\mathcal{A}}\newcommand{\M}{\mathcal{M}}\newcommand{\N}{\mathcal{N}}$I was able to fill in the blanks myself:
Let $(X,\M,\mu)$ be a $\sigma$-finite atomic space. By the linked post, we know that $X=\N\cup\bigsqcup_{n\in F}\A_n$ where $\N$ is null (and disjoint from the atoms), $F\subseteq\Bbb N$ and each $\A_n$ is an atom of finite measure disjoint from all the others. I will show both of Wikipedia's claims - there are countably many atomic classes, and if each of the atomic classes has nonempty intersection then $\mu$ is a discrete measure (a sum of Dirac measures).
The $\mu$-equivalence of class of each $\A$ is the set of all $E\in\M$ for which $\mu(\A\Delta E)=\mu(\A\setminus E)+\mu(E\setminus\A)=0$; then $\A\setminus E,E\setminus\A$ are both null.
The equivalence class of an atom consists only of atoms.
Proof:
Let $\A\in\M$, where $\A$ is an atom, and $B\in[\A]$. Then $B=(\A\cap B)\sqcup(B\setminus\A)$ and $\mu(B\Delta\A)=0$ implies that $B\setminus\A$ is null, thus $\mu(\A\cap B)=\mu(B)=\mu(A)\gt 0$ since $B$ is equivalent in measure to $\A$. Let $E\subseteq B$ be measurable. $E=(\A\cap E)\sqcup(E\setminus\A)$; $E\setminus\A\subseteq B\setminus\A$ which is null, hence $\mu(E)=\mu(\A\cap E)$. However, $\A\cap E$ is a subset of the atom $\A$ and either $\mu(E)=0$ or $\mu(E)=\mu(\A)=\mu(B)$. Hence $B$ is an atom.
Claim 1:
A $\sigma$-finite atomic space has countably many atomic classes
Proof:
Let the space be $(X,\M,\mu)$ and have the same decomposition as above. Let $\A\in\M$ be an arbitrary atom. We know that $0\lt\mu(\A)=\sum_{n\in F}\mu(\A\cap\A_n)$, since $\N$ is null, and by atomicity of $\A$, exactly one of the intersections is not null - let then $\mu(\A\cap\A_n)\gt0$ for that unique $n$. For $F\ni m\neq n$, if $\A\in[\A_m]$ then $\A\setminus\A_m$ would be null, hence $\mu(\A)=\mu(\A\cap\A_m)=0$, a contradiction. Thus $\A$ can only be in the atomic class of $[\A_n]$ or in a different one of its own. However, note that $\mu(\A\cap\A_n)\neq0\implies\mu(\A\cap\A_n)=\mu(\A)=\mu(\A_n)$ as $\A_n$ is an atom, and accordingly $\mu(\A\Delta\A_n)=0$ and $\A\in[\A_n]$. Thus the countably many atomic classes $\{[\A_n]:n\in F\}$ account for all atoms of $X$.
Now onto discreteness:
A $\sigma$-finite atomic space for which every atomic class has nonempty intersection is discrete.
Proof:
Let $(X,\M,\mu)$ and $\A_n,\N$ be as above. We have assumed: $$\forall n\in F:\exists x_n\in\bigcap_{\A\in[\A_n]}\A$$
Now let $E\in\M$ be arbitrary. $\mu(E\cap\N)=0$ as $\N$ is null, hence: $$\mu(E)=\sum_{n\in F}\mu(E\cap\A_n)$$
Let $n\in F$ be such that $E\cap\A_n$ is null. Since $\A_n=(E\cap\A_n)\sqcup(\A_n\setminus E)$ we have that $(\A_n\setminus E)\subseteq\A_n$ is not null, and inherits atomicity from $\A_n$; $\mu(\A_n\Delta(\A_n\setminus E))=\mu(\A_n\cap E)=0$ so $\A_n\setminus E\in[\A_n]$. Then by definition: $x_n\in\A_n\setminus E\in[\A_n]$ so $x_n\notin E$.
Now suppose $n\in F$ is such that $\A_n\cap E$ is not null. It follows by atomicity that $\mu(\A_n\cap E)=\mu(A),\,\mu(\A_n\setminus E)=0$ and $\A_n\cap E$ is an atom in $[\A_n]$ by the same arguments as before: then $x_n\in\A_n\cap E$ so in particular we have $x_n\in E$.
Since $x_n\in E\iff\mu(E\cap\A_n)=\mu(\A_n)\gt0$, we find that: $$\mu(E)=\sum_{n\in F:x_n\in E}\mu(\A_n)=\sum_{n\in F}\mu(\A_n)\cdot\delta_{x_n}(E)$$For all $E\in\M$. The proof is then complete, as $\mu$ is the weighted sum of Dirac measures and as such is discrete by the more conventional definitions.
Best Answer
Following Willie Wong, the answer to his question depends on the cardinality of $X$. In the case when $card(X)=c$ his question is not uniquely solvable
in the theory $(ZF)\&(DC)$ for $X=[0,1]$ and $\cal{F}=P[0,1]$, where
$(ZF)$ denotes the Zermelo-Fraenkel set theory and $(DC)$ denotes the axiom of Dependent Choices.
Indeed, on the one hand, in the consistent theory $ZF \&DC \& AD$, where $AD$ denotes an Axiom of Determinacy, the answer to his question is yes, because Mycielski and Swierczkowski well known result asserts that every subset of the real axis is Lebesgue measurable. Hence such a measure is exactly Lebesgue measure in $[0.1]$.
On the other hand, in the consistent theory $ZF\& DC \& AC \& \omega_1=2^{\omega}$ by Ulam's well known result on the powerset of $\omega_1$ (correspondingly, of $2^{\omega}$) we can not define a probability measure which vanishes on singletons.
Since both theories are consistent extensions of the theory $ZF\& DC$ we deduce that Willie Wong's question is not solvable within the theory $ZF\&DC$ for $X=[0,1]$ and $\cal{F}=P[0,1]$.