Here is a proof of this result from DC. Fix $X\in\Sigma$ such that $\mu(X)>0$ and let $a\in(0,\mu(X))$. We will use DC to inductively construct a sequence of disjoint measurable subsets $Y_n$ of $X$ such that $\mu(\bigcup Y_n)=a$.

Having defined $Y_1,\dots,Y_{n-1}$ such that $\mu(Y_1)+\dots+\mu(Y_{n-1})<a$, define $Y_n$ as follows. Let $Z=X\setminus(Y_1\cup\dots\cup Y_{n-1})$. Since $\mu$ is atomless, there exists a subset $W\subset Z$ such that $0<\mu(W)<\mu(Z)$. It is also clear that there exists such $W$ with $\mu(W)$ arbitrarily small (we can get $\mu(W)\leq \mu(Z)/2$ by replacing $W$ with $Z\setminus W$ if necessary, and now iterate); in particular, there exist such $W$ with $\mu(Y_1)+\dots+\mu(Y_{n-1})+\mu(W)<a$. Now choose $Y_n$ to be such a $W$ which additionally has the property that for any other such $W'$, $\mu(Y_n)\geq \mu(W')/2$ (that is, among the possible measures of such sets $W$, $Y_n$ is in the upper half).

Let $Y=\bigcup Y_n$. By construction, the $Y_n$ are disjoint and satisfy $\sum \mu(Y_n)\leq a$, so $\mu(Y)\leq a$. Suppose $\mu(Y)<a$. As above, there exists a subset $W\subset X\setminus Y$ such that $\mu(W)>0$ and $\mu(W)+\mu(Y)<a$. Now let $n$ be such that $\mu(Y_n)<\mu(W)/2$ (such $n$ exists since $\sum \mu(Y_n)$ is finite). Then $W$ is a subset of $X\setminus(Y_1\cup\dots\cup Y_{n-1})$ such that $\mu(Y_1)+\dots+\mu(Y_{n-1})+\mu(W)<a$, so by our choice of $Y_n$, we must have $\mu(Y_n)\geq \mu(W)/2$. This is a contradiction, so we must have $\mu(Y)=a$.

This theorem follows from Dependent Choice, and thus is strictly weaker than the Axiom of Choice. Here is a proof using only DC. Fix $X\in\Sigma$ such that $\mu(X)>0$ and let $a\in(0,\mu(X))$. We will use DC to inductively construct a sequence of disjoint measurable subsets $Y_n$ of $X$ such that $\mu(\bigcup Y_n)=a$.

Having defined $Y_1,\dots,Y_{n-1}$ such that $\mu(Y_1)+\dots+\mu(Y_{n-1})<a$, define $Y_n$ as follows. Let $Z=X\setminus(Y_1\cup\dots\cup Y_{n-1})$. Since $\mu$ is atomless, there exists a subset $W\subset Z$ such that $0<\mu(W)<\mu(Z)$. It is also clear that there exists such $W$ with $\mu(W)$ arbitrarily small (we can get $\mu(W)\leq \mu(Z)/2$ by replacing $W$ with $Z\setminus W$ if necessary, and now iterate); in particular, there exist such $W$ with $\mu(Y_1)+\dots+\mu(Y_{n-1})+\mu(W)<a$. Now choose $Y_n$ to be such a $W$ which additionally has the property that for any other such $W'$, $\mu(Y_n)\geq \mu(W')/2$ (that is, among the possible measures of such sets $W$, $Y_n$ is in the upper half).

Let $Y=\bigcup Y_n$. By construction, the $Y_n$ are disjoint and satisfy $\sum \mu(Y_n)\leq a$, so $\mu(Y)\leq a$. Suppose $\mu(Y)<a$. As above, there exists a subset $W\subset X\setminus Y$ such that $\mu(W)>0$ and $\mu(W)+\mu(Y)<a$. Now let $n$ be such that $\mu(Y_n)<\mu(W)/2$ (such $n$ exists since $\sum \mu(Y_n)$ is finite). Then $W$ is a subset of $X\setminus(Y_1\cup\dots\cup Y_{n-1})$ such that $\mu(Y_1)+\dots+\mu(Y_{n-1})+\mu(W)<a$, so by our choice of $Y_n$, we must have $\mu(Y_n)\geq \mu(W)/2$. This is a contradiction, so we must have $\mu(Y)=a$.

Here is a proof of this result from DC. Fix $X\in\Sigma$ such that $\mu(X)>0$ and let $a\in(0,\mu(X)$. We will use DC to inductively construct a sequence of disjoint measurable subsets $Y_n$ of $X$ such that $\mu(\bigcup Y_n)=a$.

Having defined $Y_1,\dots,Y_{n-1}$ such that $\mu(Y_1)+\dots+\mu(Y_{n-1})<a$, define $Y_n$ as follows. Let $Z=X\setminus(Y_1\cup\dots\cup Y_{n-1})$. Since $\mu$ is atomless, there exists a subset $W\subset Z$ such that $0<\mu(W)<\mu(Z)$. It is also clear that there exists such $W$ with $\mu(W)$ arbitrarily small (we can get $\mu(W)\leq \mu(Z)/2$ by replacing $W$ with $Z\setminus W$ if necessary, and now iterate); in particular, there exist such $W$ with $\mu(Y_1)+\dots+\mu(Y_{n-1})+\mu(W)<a$. Now choose $Y_n$ to be such a $W$ which additionally has the property that for any other such $W'$, $\mu(Y_n)\geq \mu(W')/2$ (that is, among the possible measures of such sets $W$, $Y_n$ is in the upper half).

Let $Y=\bigcup Y_n$. By construction, the $Y_n$ are disjoint and satisfy $\sum \mu(Y_n)\leq a$, so $\mu(Y)\leq a$. Suppose $\mu(Y)<a$. As above, there exists a subset $W\subset X\setminus Y$ such that $\mu(W)>0$ and $\mu(W)+\mu(Y)<a$. Now let $n$ be such that $\mu(Y_n)<\mu(W)/2$ (such $n$ exists since $\sum \mu(Y_n)$ is finite). Then $W$ is a subset of $X\setminus(Y_1\cup\dots\cup Y_{n-1})$ such that $\mu(Y_1)+\dots+\mu(Y_{n-1})+\mu(W)<a$, so by our choice of $Y_n$, we must have $\mu(Y_n)\geq \mu(W)/2$. This is a contradiction, so we must have $\mu(Y)=a$.

Here is a proof of this result from DC. Fix $X\in\Sigma$ such that $\mu(X)>0$ and let $a\in(0,\mu(X))$. We will use DC to inductively construct a sequence of disjoint measurable subsets $Y_n$ of $X$ such that $\mu(\bigcup Y_n)=a$.

Having defined $Y_1,\dots,Y_{n-1}$ such that $\mu(Y_1)+\dots+\mu(Y_{n-1})<a$, define $Y_n$ as follows. Let $Z=X\setminus(Y_1\cup\dots\cup Y_{n-1})$. Since $\mu$ is atomless, there exists a subset $W\subset Z$ such that $0<\mu(W)<\mu(Z)$. It is also clear that there exists such $W$ with $\mu(W)$ arbitrarily small (we can get $\mu(W)\leq \mu(Z)/2$ by replacing $W$ with $Z\setminus W$ if necessary, and now iterate); in particular, there exist such $W$ with $\mu(Y_1)+\dots+\mu(Y_{n-1})+\mu(W)<a$. Now choose $Y_n$ to be such a $W$ which additionally has the property that for any other such $W'$, $\mu(Y_n)\geq \mu(W')/2$ (that is, among the possible measures of such sets $W$, $Y_n$ is in the upper half).

Let $Y=\bigcup Y_n$. By construction, the $Y_n$ are disjoint and satisfy $\sum \mu(Y_n)\leq a$, so $\mu(Y)\leq a$. Suppose $\mu(Y)<a$. As above, there exists a subset $W\subset X\setminus Y$ such that $\mu(W)>0$ and $\mu(W)+\mu(Y)<a$. Now let $n$ be such that $\mu(Y_n)<\mu(W)/2$ (such $n$ exists since $\sum \mu(Y_n)$ is finite). Then $W$ is a subset of $X\setminus(Y_1\cup\dots\cup Y_{n-1})$ such that $\mu(Y_1)+\dots+\mu(Y_{n-1})+\mu(W)<a$, so by our choice of $Y_n$, we must have $\mu(Y_n)\geq \mu(W)/2$. This is a contradiction, so we must have $\mu(Y)=a$.