= Solution
For each integer $m\geq1$, let
$$
D_m=\{x\in\mathbb R:\mu(\{x\})\geq1/m\}.
$$
If $D_m$ had $m+1$ distinct members, <finite additivity> on their disjoint singleton <Borel sets> would give total <measure> at least $(m+1)/m>1$, contradicting $\mu(\mathbb R)=1$. Thus $|D_m|\leq m$.
Every positive singleton <measure> is at least $1/m$ for some sufficiently large $m$. Consequently the set of <point atoms of a measure> is
$$
\{x:\mu(\{x\})>0\}=\bigcup_{m\geq1}D_m.
$$
A <countable union> of finite <sets> is at most <countable>. This proves the claim, including the possibility that the set is empty. The same argument gives the <countability of point atoms of a finite measure>, with $|D_m|\leq m\mu(\mathbb R)$ in place of $m$.
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