If $|X|$ is a [measurable cardinal][1], then there is a $\sigma$-complete ultrafilter $\mathcal U$ on $X$. You may define a countably additive probability measure $\mu$ on $(X,2^X)$ by setting $\mu(Y) = 1$ if $Y \in \mathcal U$ and $\mu(Y) = 0$ if $Y \notin \mathcal U$. This measure takes only two values (hence is atomic) and assigns measure $0$ to every singleton. Thus the answer to your question is "no it does not follow" (assuming the existence of a measurable cardinal).

Furthermore, we can show that the measurable cardinal is necessary, in the sense that if there is an example $\mu$ of an atomic measure having the properties you describe, then there is a measurable cardinal. To see this, first note that any such measure $\mu$ must be atomic, by the theorem you quoted in your post. Fix $Y \subseteq X$ such that $\mu(Y) = c > 0$ and if $Z \subseteq Y$ then either $\mu(Z) = 0$ or $\mu(Z) = c$. (This is what it means for a measure to be [atomic][2].) Letting $\mathcal U = \{Z \subseteq Y : \mu(Z) = c\}$, it is not difficult to show that $\mathcal U$ is a $\sigma$-complete ultrafilter on $Y$. This shows that $|Y|$ is $\geq$ the least measurable cardinal.

Thus a fuller answer to your question is "no it doesn't follow . . . or if it does, then the existence of measurable cardinals is inconsistent."


  [1]: http://cantorsattic.info/Measurable
  [2]: https://en.wikipedia.org/wiki/Atom_(measure_theory)