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.
The statements follow from the definitions of different convergence properties.
For weak convergence, we need to check
$$\lim_{n\to \infty}\int_{\mathbb{R}} \phi \ d \delta_{\frac 1 n} =\int_{\mathbb{R}} \phi \ d \delta_{0},$$
for any given bounded continuous function $\phi$ on $\mathbb{R}$. This is true since by definition of Dirac measure and continuity of $\phi$, we have
$$\lim_{n\to \infty}\int_{\mathbb{R}} \phi \ d \delta_{\frac 1 n} =\lim_{n\to \infty} \phi({\frac 1 n})= \phi(0) =\int_{\mathbb{R}} \phi \ d \delta_{0}.$$
For strong convergence, we need to check $$\lim_{n\to\infty} \delta_{\frac 1 n} (A)= \delta_{0} (A),$$
for any given Borel measurable set $A\subset \mathbb{R}$. This is false since if we take $A=(0,1)$, then
$ \delta_{\frac 1 n} (A)=1$ for all $n$ but $\delta_{0} (A)=0$.
Convergence in TV norm is even stronger than the strong convergence, since we need to check $$\lim_{n\to\infty}\sup_{A} |\delta_{\frac 1 n} (A)-\delta_{0} (A)|=0,$$
where the supremum is taken over all Borel measurable sets $A\subset \mathbb{R}$. The above counterexample shows that it is false.
Best Answer
First we should note that "continuous" is not the same as "non-atomic". By definition $\mu$ is continuous if $\mu(\{x\})=0$ for every $x$, while $\mu$ is non-atomic if there are no atoms, where $E$ is an atom if $\mu(E)>0$ and every (measurable) subset of $E$ has measure $0$ or $\mu(E)$. So a non--atomic measure must be continuous, but for example the Dieudonne measure on $\omega_1$ is continuous but not non-atomic.
In any case the answer to your question is no. Say $(x_n)$ and $\mu$ are as in the question. Suppose $y_n\to y$ and let $(z_n)$ be the sequence $$x_1,y_1,x_2,y_2,\dots.$$Then $(z_n)$ is dense, but$$\frac1n\sum_{i=1}^n\delta_{z_i}\to\frac12(\mu+\delta_y).$$
New Question, regarding continuous measures on compact metric spaces: If $X$ is a countable compact metric space then there is no continuous probability measure. If $X$ also has no isolated points then of course there is. One could construct a continuous probability measure of the form $\lim\frac1n\sum_1^n\delta_{x_i}$, but the details involved in ensuring that $\mu$ is continuous would be, at least in the construction I have in mind, more or less a copy of the proof that $X$ contains a Cantor set.
Edit: Just saw a nice clean way to show there is a continuous Borel probability measure on a compact metric space with no isolated points. Not exactly of the form $\lim\frac1n\sum_1^n\delta_{x_i}$, but in the same spirit: It is the limit of averages of point masses:
Since $X$ has no isolated points you can define closed balls $B_\alpha=\overline{B(x_\alpha,r_\alpha)}$ for each finite sequence of zeroes and ones $\alpha$ such that $$B_{\alpha,0},B_{\alpha,1}\subset B_\alpha$$and $$B_{\alpha,0}\cap B_{\alpha,1}=\emptyset.$$ Letting $|\alpha|$ denote the length of $\alpha$, define $$\mu_n=2^{-n}\sum_{|\alpha|=n}\delta_{x_\alpha}.$$
If you assume in addition that $r_\alpha\to0$ as $|\alpha|\to\infty$ then $\mu=\lim\mu_n$ exists; if you don't want to worry about that, just let $\mu$ be a weak-* accumulation point of $(\mu_n)$.
Now since $B_\alpha$ is closed it follows that $$\mu(B_\alpha)\ge 2^{-|\alpha|},$$and since $\mu(X)=1$ this shows that in fact $$\mu(B_\alpha)=2^{-|\alpha|}.$$Hence $$\lim_{r\to0}\mu(B(x,r))=0$$for every $x$, so $\mu$ is continuous. (Let $K=\bigcap_n\bigcup_{|\alpha|=n}B_\alpha$, and consider the cases $x\in K$ and $x\notin K$ separately...)