An equivalent condition is that the probability space $(X, \mathcal{M}, \mu)$ is non-atomic, meaning that for every   $A\in \mathcal{M}$  with $\mu(A)>0$ there exists $B \in \mathcal{M}$ such that $B\subset A$ and $0<\mu(B)<\mu(A)$.

If  $(X, \mathcal{M}, \mu)$ is non-atomic then $\mu: \mathcal{M}\to [0,1]$ has a monotone section: that is, a map $E:[0,1]\to \mathcal{M}$ such that for all  $0\le t\le s\le1$ one has $\emptyset=E_0\subset E_t\subset E_s\subset E_1=X$  and  $\mu(E_t)=t$. 

In fact, any "partial monotone section of $\mu$" (meaning a family $E:S\to \mathcal{M}$ as above, but with possibly smaller domain $S\subset [0,1]$) can be extended to a monotone section of $\mu$ (which in particular solves your problem). 

I think this result is essentially due to Sierpiński; the proof uses the Zorn lemma on the set of graphs of the partial monotone sections of $\mu$, ordered by "extension" i.e. inclusion of graphs. 

By the way, the maximality argument via Zorn lemma generalizes to a proof of the Lyapunov convexity theorem: if a $\mathbb{R}^n$ valued bounded vector measure  $\mu=(\mu_1,\dots,\mu_n)$ is non atomic (meaning, all $\mu_i$ are non-atomic bounded measures), then the range of $\mu$ is a compact convex set.