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Short version: the set $\{\mu(B):B\in\mathcal{B}\}$ is a closed set for any probability space $(X,\mathcal{B},\mu)$. For atomic spaces this follows from an elementary topological argument, and for non-atomic spaces it is a closed interval by a classical (and easy) result of Sierpinski.

Longer version with details:

Let $A_n=A'_n\cup A''_n$ where $A'_n$ is the atomic part of $A_n$ and $A''_n$ is the rest. By passing to a subsequence we may assume that $\mu(A'_n)$ converges to some $\alpha\in [0,1/2]$.

I claim that there is an atomic set $B'$ such that $\mu(B')=\alpha$. Indeed, let $\{ x_k\}$ be the collection of the $\mu$-measures of all atoms of all $A'_n$.

The set $X=\{ \sum_{k\in I} x_k: I\subset \mathbb{N}\}$ can be checked to be closed, since it is the continuous image of the $\{0,1\}^\mathbb{N}$ under the map $I\to \sum_{k\in I} x_k$, where we identify sequences $\omega$ in $\{0,1\}^\mathbb{N}$ with the set $I=\{ j: \omega_j=1\}$. Since $\alpha$ is in the closure of $X$ (as $\mu(A'_n)\to\alpha$), $\alpha\in X$, as claimed.

Now the non-atomic part of the space has mass at least $1/2-\alpha$ (since it has mass at least $\mu(A''_n)$ for each $n$). But it is well known that if $\nu$ is a non-atomic measure of mass $c$, then for any $c'\in [0,c]$ there is a measurable set of $\nu$-mass $c'$ (see for example wikipedia). Applying this to $\nu$=restriction of $\mu$ to non-atomic part and $c'=1/2-\alpha$, we get a measurable non-atomic set $B''$ of $\mu$-measure $1/2-\alpha$.

Hence $B=B'\cup B''$ satifies $\mu(B)=1/2$.

Short version: the set $\{\mu(B):B\in\mathcal{B}\}$ is a closed set for any probability space $(X,\mathcal{B},\mu)$. For atomic spaces this follows from an elementary topological argument, and for non-atomic spaces it is a closed interval by a classical (and easy) result of Sierpinski.

Longer version with details:

Let $A_n=A'_n\cup A''_n$ where $A'_n$ is the atomic part of $A_n$ and $A''_n$ is the rest. By passing to a subsequence we may assume that $\mu(A'_n)$ converges to some $\alpha\in [0,1/2]$.

I claim that there is an atomic set $B'$ such that $\mu(B')=\alpha$. Indeed, let $\{ x_k\}$ be the collection of the $\mu$-measures of all atoms of all $A'_n$.

The set $X=\{ \sum_{k\in I} x_k: I\subset \mathbb{N}\}$ can be checked to be closed, since it is the continuous image of the $\{0,1\}^\mathbb{N}$ under the map $I\to \sum_{k\in I} x_k$, where we identify sequences $\omega$ in $\{0,1\}^\mathbb{N}$ with the set $I=\{ j: \omega_j=1\}$. Since $\alpha$ is in the closure of $X$ (as $\mu(A'_n)\to\alpha$), $\alpha\in X$, as claimed.

Now the non-atomic part of the space has mass at least $1/2-\alpha$ (since it has mass at least $\mu(A''_n)$ for each $n$). But it is well known that if $\nu$ is a non-atomic measure of mass $c$, then for any $c'\in [0,c]$ there is a measurable set of $\nu$-mass $c'$ (see for example Wikipedia). Applying this to $\nu$=restriction of $\mu$ to non-atomic part and $c'=1/2-\alpha$, we get a measurable non-atomic set $B''$ of $\mu$-measure $1/2-\alpha$.

Hence $B=B'\cup B''$ satifies $\mu(B)=1/2$.

Let $A_n=A'_n\cup A''_n$ where $A'_n$ is the atomic part of $A_n$ and $A''_n$ is the rest. By passing to a subsequence we may assume that $\mu(A'_n)$ converges to some $\alpha\in [0,1/2]$.

I claim that there is an atomic set $B'$ such that $\mu(B')=\alpha$. Indeed, let $\{ x_k\}$ be the collection of the $\mu$-measures of all atoms of all $A'_n$.

The set $X=\{ \sum_{k\in I} x_k: I\subset \mathbb{N}\}$ can be checked to be closed, since it is the continuous image of the $\{0,1\}^\mathbb{N}$ under the map $I\to \sum_{k\in I} x_k$, where we identify sequences $\omega$ in $\{0,1\}^\mathbb{N}$ with the set $I=\{ j: \omega_j=1\}$. Since $\alpha$ is in the closure of $X$ (as $\mu(A'_n)\to\alpha$), $\alpha\in X$, as claimed.

Now the non-atomic part of the space has mass at least $1/2-\alpha$ (since it has mass at least $\mu(A''_n)$ for each $n$). But it is well known that if $\nu$ is a non-atomic measure of mass $c$, then for any $c'\in [0,c]$ there is a measurable set of $\nu$-mass $c'$ (see for example wikipedia). Applying this to $\nu$=restriction of $\mu$ to non-atomic part and $c'=1/2-\alpha$, we get a measurable non-atomic set $B''$ of $\mu$-measure $1/2-\alpha$.

Hence $B=B'\cup B''$ satifies $\mu(B)=1/2$.

Short version: the set $\{\mu(B):B\in\mathcal{B}\}$ is a closed set for any probability space $(X,\mathcal{B},\mu)$. For atomic spaces this follows from an elementary topological argument, and for non-atomic spaces it is a closed interval by a classical (and easy) result of Sierpinski.

Longer version with details:

Let $A_n=A'_n\cup A''_n$ where $A'_n$ is the atomic part of $A_n$ and $A''_n$ is the rest. By passing to a subsequence we may assume that $\mu(A'_n)$ converges to some $\alpha\in [0,1/2]$.

I claim that there is an atomic set $B'$ such that $\mu(B')=\alpha$. Indeed, let $\{ x_k\}$ be the collection of the $\mu$-measures of all atoms of all $A'_n$.

The set $X=\{ \sum_{k\in I} x_k: I\subset \mathbb{N}\}$ can be checked to be closed, since it is the continuous image of the $\{0,1\}^\mathbb{N}$ under the map $I\to \sum_{k\in I} x_k$, where we identify sequences $\omega$ in $\{0,1\}^\mathbb{N}$ with the set $I=\{ j: \omega_j=1\}$. Since $\alpha$ is in the closure of $X$ (as $\mu(A'_n)\to\alpha$), $\alpha\in X$, as claimed.

Now the non-atomic part of the space has mass at least $1/2-\alpha$ (since it has mass at least $\mu(A''_n)$ for each $n$). But it is well known that if $\nu$ is a non-atomic measure of mass $c$, then for any $c'\in [0,c]$ there is a measurable set of $\nu$-mass $c'$ (see for example wikipedia). Applying this to $\nu$=restriction of $\mu$ to non-atomic part and $c'=1/2-\alpha$, we get a measurable non-atomic set $B''$ of $\mu$-measure $1/2-\alpha$.

Hence $B=B'\cup B''$ satifies $\mu(B)=1/2$.

Let $A_n=A'_n\cup A''_n$ where $A'_n$ is the atomic part of $A_n$ and $A''_n$ is the rest. By passing to a subsequence we may assume that $\mu(A'_n)$ converges to some $\alpha\in [0,1/2]$.

I claim that there is an atomic set $B'$ such that $\mu(B')=\alpha$. Indeed, if we let $\{ x_k\}$ be collection of the $\mu$-measures of all atoms of all $A'_n$, we know that $\alpha\le \sum_k x_k \le 1$.

Now the set $X=\{ \sum_{k\in I} x_k: I\subset \mathbb{N}\}$ can be checked to be closed, since it is the continuous image of the $\{0,1\}^\mathbb{N}$ under the map $I\to \sum_{k\in I} x_k$, where we identify sequences $\omega$ in $\{0,1\}^\mathbb{N}$ with the set $I=\{ j: \omega_j=1\}$. Since $\alpha$ is in the closure of $X$ (as $\mu(A'_n)\to\alpha$), $\alpha\in X$, as claimed.

Now the non-atomic part of the space has mass at least $1/2-\alpha$ (since it has mass at least $\mu(A''_n)$ for each $n$). But it is well known that if $\nu$ is a non-atomic measure of mass $c$, then for any $c'\in [0,c]$ there is a measurable set of $\nu$-mass $c'$ (see for example wikipedia). Applying this to $\nu$=restriction of $\mu$ to non-atomic part and $c'=1/2-\alpha$, we get a measurable non-atomic set $B''$ of $\mu$-measure $1/2-\alpha$.

Hence $B=B'\cup B''$ satifies $\mu(B)=1/2$.

Let $A_n=A'_n\cup A''_n$ where $A'_n$ is the atomic part of $A_n$ and $A''_n$ is the rest. By passing to a subsequence we may assume that $\mu(A'_n)$ converges to some $\alpha\in [0,1/2]$.

I claim that there is an atomic set $B'$ such that $\mu(B')=\alpha$. Indeed, let $\{ x_k\}$ be the collection of the $\mu$-measures of all atoms of all $A'_n$.

The set $X=\{ \sum_{k\in I} x_k: I\subset \mathbb{N}\}$ can be checked to be closed, since it is the continuous image of the $\{0,1\}^\mathbb{N}$ under the map $I\to \sum_{k\in I} x_k$, where we identify sequences $\omega$ in $\{0,1\}^\mathbb{N}$ with the set $I=\{ j: \omega_j=1\}$. Since $\alpha$ is in the closure of $X$ (as $\mu(A'_n)\to\alpha$), $\alpha\in X$, as claimed.

Now the non-atomic part of the space has mass at least $1/2-\alpha$ (since it has mass at least $\mu(A''_n)$ for each $n$). But it is well known that if $\nu$ is a non-atomic measure of mass $c$, then for any $c'\in [0,c]$ there is a measurable set of $\nu$-mass $c'$ (see for example wikipedia). Applying this to $\nu$=restriction of $\mu$ to non-atomic part and $c'=1/2-\alpha$, we get a measurable non-atomic set $B''$ of $\mu$-measure $1/2-\alpha$.

Hence $B=B'\cup B''$ satifies $\mu(B)=1/2$.