Let $X$ be a separable metric space and $\mathcal{F}$ be the borel $\sigma$-algebra of $X$. Let $\mu$ be a $\sigma$-finite atomic or purely atomic measure on $\mathcal{F}$ i.e. every measurable set of positive measure contains an atom. This is equivalent to say that there is a countable partition of $X$ formed by atoms up to a null set.
Let $E$ be an atom such that $\mu(E)<\infty$. Since $X$ is separable metric space, $X$ can be covered by countably many closed balls of radius $1/2$ say $\{B_i\}$. Then $E=\bigcup\limits_{i=1}^\infty E\cap B_i\implies \mu(E)\leq\sum\limits_{i=1}^\infty\mu(E\cap B_i)$. But each $\mu(E\cap B_i)$ is either $0$ or $\mu(E)$. So there is $i\in\Bbb{N}$ such that $\mu(E\cap B_i)=\mu(E)$ and $\mu(E\cap B_j)=0\ \forall i\ne j$. We call $E_1=E\cap B_i$. Then $\mu(E_1)=\mu(E)$ with $\text{diam}(E_1)\le1$. Similarly as $E_1$ is atom, there is $E_2\subseteq E_1$ such that $\mu(E_2)=\mu(E_1)=\mu(E)$ such that $\text{diam}(E_2)\le 1/2$. Inductively, there is a decreasing sequence of atoms $\{E_n\}$ such that $\mu(E_n)=\mu(E)$ and $\text{diam}(E_n)\le 1/n$.
Now can I conclude $\bigcap\limits_{n=1}^\infty E_n=\{x\}$? If so, then $\mu(x)=\lim\mu(E_n)=\mu(E)$, therefore every atom of finite measure differs from a singleton by a null set.
Now as $\mu$ is atomic, $X$ can be partitioned into countably many atoms upto a null set. WLOG, assume $X=\bigsqcup\limits_{i=1}^\infty A_i$ where each $A_i$ is atom. As $\mu$ is $\sigma$-finite $\mu(A_i)<\infty\ \forall i$. So for each $i$, there is $x_i\in A_i$ such that $\mu(x_i)=\mu(A_i)$. We claim that $\mu$ is supported by $\{x_i:\ i\in\Bbb{N}\}$. If not, then $\mu(X\setminus\{x_i\})>0$, hence it contains an atom say $E$. Thereofore $x_i\notin E\ \forall i$. Now $E=\bigsqcup\limits_{i=1}^\infty E\cap A_i\implies \mu(E)=\sum\limits_{i=1}^\infty \mu(E\cap A_i)$. Since $E$ is atom, only one of the members in that sum is non-zero. WLOG, assume $\mu(E\cap A_1)>0$. Again, as $A_1$ is atom, $\mu(E\cap A_1)=\mu(A_1)$. Now $\mu(x_1)+\mu(E\cap A_1)\le \mu(A_1)\implies 2\mu(A_1)\le\mu(A_1)$, this is a contradiction. So, we are done.
So my actual problem where this proof is stuck or may not hold is- $\bigcap\limits_{n=1}^\infty E_n=\{x\}$?
Can anyone help to complete my proof? Or, if this does not hold any idea to improve this? Thanks for your help in advance.