Let $(X, d)$ be a compact metric space. We denote by $\mathcal{M}(X)$ the set of all Borel probability measures of $X$. Also, $\mu_n\to \mu$ if and only if $\int f d\mu_n\to \int f d\mu$ for every continuous map $f:X\to \mathbb{R}$.
Denote by $m_x$ the Dirac measure supported on $x\in X$, indeed, $m_x(A)=0$ or $1$ depending on whether $x\notin A$ or $x\in A$.
We have a sequence $\{x_n\}_{n\in\mathbb{N}}\subseteq X$ with $x_n\to x$. What can say about $\{m_{x_n}\}_{n\in\mathbb{N}}$. Is it true that $m_{x_n}\to m_x$ as $n\to \infty$?
The answer to the last question is yes. The Lemma and the Theorem below summarizes the properties of embedding the space into its space of measures.
We also point out that the statement below hold not only in the setting of the OP but in the slightly more general setting of separable and complete metric spaces (Polish spaces) that is used extensively in the Theory of Probability.
Let $(S,d)$ be a Polish space and denote by $\mathscr{B}(S)$ the Borel $\sigma$-algebra, and let $\mathcal{M}(S)$ be the space of all finite Borel measures on $\mathscr{B}(S)$. Denote by $\mathcal{C}_b(S)$ the space of all bounded continuous functions on $S$. The weak topology $\sigma(\mathcal{M}(S),\mathcal{B}_b(S))$ is the smallest topology on $\mathcal{M}(S)$ that makes the the functions $$\Lambda_f:\mu\mapsto \int_S f\,d\mu$$ for each $f\in\mathcal{C}_b(S)$ continuous. A local basis for this topology is generated by sets of the form $$V(f_1,\ldots,f_n;\varepsilon)=\{|\Lambda_{f_j}\mu|<\varepsilon,\,1\leq j\leq n\}$$ where $n\in\mathbb{Z}_+$, $f_1,\ldots,f_n\in\mathcal{C}_b(S)$, and $\varepsilon>0$.
Observation If $(S,d)$ is a compact metric space then $\mathcal{B}_b(S)=\mathcal{C}_{00}(S)=\mathcal{C}(S)$, where $\mathcal{C}_{00}(S)$ denotes the space of continuous functions on $S$ with compact support, and $\mathcal{C}(S)$ denotes the space of continuous functions on $S$. By the Riesz representation theorem $(\mathcal{C}(S),\|\;\|_u)^*=(\mathcal{M}(S),\|\;\|_{TV})$ where $\|\;\|_{TV}$ is the total variotion norm. In this case, the weak-* topology on $\mathcal{M}(S)$ coincides with the weak-topology defines above.
The space $S$ itself embeds into $\big(\mathcal{M}(S),\sigma(\mathcal{M}(S),\mathcal{C}_b(S))\big)$ through the map $x\mapsto\delta_x$. The next results show that this mapping is continuous and that $S$ (the embeding) is a weakly closed.
For each $x\in S$, denote by $\delta_x$ the measure in $\mathcal{M}_+(S)$ such that $\delta_x(A)=\mathbb{1}_A(x)$ for all $A\in\mathscr{B}(S)$.
Proof: $\delta_{x_\alpha}\Rightarrow\delta_x$ iff $\lim_\alpha f(x_\alpha)= f(x)$ for all $f\in\mathcal{C}_b(S)$. The particular choice $f(y)=1\wedge d(y,x)$ implies that $x_\alpha\rightarrow x$. $\Box$
Let $\mathcal{M}_+(S)\subset\mathcal{M}(S)$ be the subspace nonnegative finite Borel measures on $\mathscr{B}(S)$. The following result summarizes the properties of the embedding of $S$ into $\mathcal{M}_+(S)$
Proof: Suppose that $\delta_{x_\alpha}\Rightarrow \mu$. As $\mathcal{M}_+(S)$ is a closed in $\big(\mathcal{M}(S),\sigma(\mathcal{M}(S),\mathcal{C}_b(S))\big)$ and $\{\delta_x:x\in s\}\subset\mathcal{M}_+(S)$, $\mu\in\mathcal{M}_+(S)$.
Let $x\in\operatorname{supp}(\mu)$. For any open neighborhood $V$ of $x$, let $f\in\mathcal{C}_b(S)$ be such that $0\leq f\leq 1$, $f(x)=1$ and $f=0$ on $ S\setminus V$. Then $\int f\,d\mu>0$ and, as $$\lim_\alpha\delta_{x_\alpha}f=\lim_\alpha f(x_\alpha)= \mu f,$$ there exists $\alpha_0\in D$ such that $\alpha\geq\alpha_0$ implies that $x_\alpha\in V$. Hence, $x_\alpha\rightarrow x$, and by the Lemma presented above, $\mu=\delta_x$. $\Box$