As note 'added in proof': the above answers in the positive the point I was trying to make, i.e. that in the presence of a gap at some $r>0$, the measure $\mu$ cannot be atomless.

Indeed, suppose that $\mu$ has a gap at $r$. Let $\alpha=\sup\{\mu(A):A\in \cal C\}$ and $\beta=\inf\{\mu(A):A\in \cal M, \mu(A)\ge r\}$. If $\mu$ were atomless, we know that for any $\epsilon>0$, we could write $X=\cup_{n\ge 1}A_n$ where the $A_n$'s are measurable sets of measure $\le \epsilon$. Hence for some $N\ge 1$ we would have $\alpha\le \mu(\cup_{n=1}^{N} A_n)< \alpha+\epsilon$, and taking $\epsilon<\beta-\alpha$ would yield a contradiction. Hence $\mu$ must have an atom.

As note 'added in proof': the above answers in the positive the point I was trying to make, i.e. that in the presence of a gap at some $r>0$, the measure $\mu$ cannot be atomless.

Indeed, suppose that $\mu$ has a gap at $r$. Let $\alpha=\sup\{\mu(A):A\in \cal C\}$ and $\beta=\inf\{\mu(A):A\in \cal M, \mu(A)\ge r\}$. If $\mu$ were atomless, we know that for any $\epsilon>0$, we could write $X=\cup_{n\ge 1}A_n$ where the $A_n$'s are measurable sets of measure $\le \epsilon$. Hence for some $N\ge 1$ we would have $\alpha< \mu(\cup_{n=1}^{N} A_n)\le \alpha+\epsilon$, and taking $\epsilon<\beta-\alpha$ would yield a contradiction. Hence $\mu$ must have an atom.