Without loss of generality your atom $A$ is compact (by inner regularity). Call a point $a\in A$ negligible if it has a neighborhood with zero measure inside $A$. Clearly the set $B$ of negligible points is open in $A$. If it is $A$ itself, then choose finite subcovering by those neighborhoods to get that measure of $A$ is zero. So, $A\setminus B=C$ is a non-empty compact set in $A$. It has either measure 0 or measure equal to $|A|$ (let $|\cdot|$ denote measure.) If $|C|=0$, then choose a neighborhood $V$ of $C$ with measure at most $|A|/2$ by outer regularity. $V\cap A$ has measure strictly less then $|A|$, hence 0, hence each point of $C$ is negligible. A contradiction. So $|C|=|A|$. Then replace $A$ for $C$ and we get an atom $C$ in each no point is negligible. It may contain only one point, else take two points $u$, $v$ and their disjoint neighborhoods. Both must have positive measure in $C$, a contradiction.