(1) Existence of sets of small measure.

Let $(\Omega,\Sigma,\mu)$ an atomless finite measure space with $\mu(\Omega)>0$.
It is easy to show that any measurable set $A$ with $\mu(A)>0$ contains a measurable subset $B\subset A$ with $0<\mu(B)\le \mu(A)/2$, and therefore a measurable set $M$ of measure $0<\mu(M)\le \varepsilon$. 

(2) The space and any measurable set is a numerable union of set of small measure.

Now given $\varepsilon>0$ consider the set $\mathcal M$ of measurable sets $M=\bigcup_n A_n$
union of a numerable set of measurable sets of measure $\mu(A_n)\le \varepsilon$

Let $\lambda$ the supremum of the measures $\mu(M)$ of set in $\mathcal M$.
There is a sequence $(M_n)$ of sets in $\mathcal M$ with $\lim_n\mu(M_n)=\lambda$. Then $M_0:=\bigcup M_n\in \mathcal M$ (by definition) and $\lambda(M_0)=\lambda$. But it follows that $\lambda=\mu(\Omega)$, because in other case $\Omega\smallsetminus M_0$ will contain a set of measure $\le \varepsilon$, which contradict the maximality of $M_0$. 

(3) Given a measurable set $M$ and two numbers $0<a<b<\mu(M)$, there exists a measurable set $A\subset M$ with $a<\mu(A)<b$. 

Let $\varepsilon=(b-a)/2$, by (2) the set $M$ can be put as a union of measurable sets of measure less than $\varepsilon$,  $M=\bigcup_n M_n$ with $\mu(M_n)\le \varepsilon$.  Then $$\lim_N\mu\Bigl(\bigcup_{n=1}^N M_n\Bigr)=\mu(M).$$ Therefore there is some $N$ such that $a<\mu\Bigl(\bigcup_{n=1}^N M_n\Bigr)<b$, since each $M_n$ have measure $\le (b-a)/2$. 

(4) Construct the set with measure $\mu(A)=a$ for a given $0<a<\mu(\Omega)$.

We will construct a sequence of measurable sets $(A_n)$ with $A_n\le A_{n+1}$ and such that 
$$\frac{a+\mu(A_n)}{2}<\mu(A_{n+1})< a.$$
It is clear that in this case $A=\bigcup_n A_n$ have measure $a$.

Start with $A_0=\emptyset$. We must construct $A_1$ with $a/2<\mu(A_1)<a$. Sincer $\Omega$ have measure $>a$, we have constructed such a set in (3). 

Assume we have constructed $A_n$, then $\mu(A_n)<a$, applying (3) we construct $B_n\subset \Omega\smallsetminus A_n$ whose measure is greater than half the difference with our objective:  $a-\mu(A_n)>\mu(B_n)>\frac{a-\mu(A_n)}{2}$. 
Hence  $A_{n+1}=A_n\cup B_n$ satisfies our requirement. First $A_n$ and $B_n$ are disjoint $\mu(A_{n+1})=\mu(A_n)+\mu(B_n)<a$, and $\mu(A_n)+\mu(B_n)>\frac{a+\mu(A_n)}{2}$.