Show that there is $E \in M$ such that $\mu(E)=\alpha$ for any $0 \lt \alpha \lt \mu(X)$

Let $(X,M,\mu)$ be a finite measure space with no atoms. A set $A \in M$ is called an atom if $\mu(A) \gt 0$ and for any measurable subset $B \subset A$ ,either $\mu(B)=0$ or $\mu(A-B)=0$ .Show that there is $E \in M$ such that $\mu(E)=\alpha$ for any $0 \lt \alpha \lt \mu(X)$.

My attempt:

For any $F \in M$, there exists a $B \in F$ such that $\mu(B) \ne 0$ and $\mu(F-B) \ne0$ (Measure Space has no atoms).Since $\mu$ is finite, there exists a $k_0 \in \mathbb{N}$ such that $\mu(F) \lt k_0$. Then since $\mu(F)=\mu(B)+\mu(F-B)$. Atleast one of $B$ or $F-B$ has measure less than $\dfrac{\mu(F)}{2}$. Call it $B_1$. Similarly we can find $B_2 \subset B_1$ such that $\mu(B_2) \lt \dfrac{\mu(F)}{2^2}$. By induction we can find $B_n$ such that $B_n \subset B_{n-1}\subset....\subset B_2 \subset B_1$ such that $\mu(B_n)\lt \dfrac{\mu(F)}{2^n}$. This just means that given $\epsilon \gt 0$ , there is a set $S \in M$ such that $\mu(S) \lt \epsilon$.

Now I try to define a set $$T=\{E \in M| \mu(E) \le \alpha\}$$. This set is nonempty by the above observation. I define the order relation $"\le"$ by inclusion. Then $T$ is a partially ordered set. Any linearly ordered set $L$ in $T$ has sets which are nested. The upper bound is the union of all these, which is unfortunately not in $M$, for the union can be be over any index, countable and uncountable.

This makes me to change the set to $$T'=\{E_{\alpha} \in M| \mu(E_{\alpha}) \le \alpha,\mu(E_{\alpha}\cap E_{\beta})=\phi, \text{for} ,\alpha, \beta \in I, \alpha \ne \beta\}$$

Then since $\mu(X)\lt \infty$, there are atmost countably many of these sets as defined in $T'$. But I am unable to find an order relation for this set. Another way could be to define an order relation such that any linearly ordered set has of the form $T'$.

I tried another possible approach:Define ~ on $ M $ by $ E $~ $F$ iff $\mu ((E\cap F^{c}) \cup (E^{c}\cap F^{c}))=0$. Then this is an equivalence relation on $ M $ and defining $ d ([E], [F])= \mu ((E\cap F^{c}) \cup (E^{c}\cap F^{c})$ turns $M $ into a metric space. Now the problem just boils down to showing that the space $( M$,~$) $ is connected with the above metric. Thanks for the help!!