Prove that if $0
Suppose $\Pr$ is non-atomic that is $\Pr(A)$ implies the existence of $B$ with $B\subset A$ such that $0<\Pr(B)<\Pr(A) $ and $A\in\mathcal{F}$ with $\Pr(A)>0$.
(a)show that for every $\epsilon>0$, we have $B\subset A$ such that $0<\Pr(B)<\epsilon$.
(b)Prove that if $0<a<\Pr(A)$ then there exists $B\subset A$ such that $\Pr(B)=a$.
Here is a hint:
fix $\epsilon_n \to 0$ and define inductively numbers $x_n$ and sets $G\in \mathcal{F}$ with $H_0=\emptyset$, $H_n=\cup_{k<n} G_k$, $x_n=\sup\{\Pr(G): G\subset A\setminus H_n, \Pr(H_n\cup G)\leq a\}$ and $G_n\subset A\setminus H_n$
such that $\Pr(H_n\cup G_n)\leq a$ and $\Pr(G_n)\geq (1-\epsilon_n)x_n$. Let $B=\cup_{k} G_k$.
 A: It suffices to prove for the second question, for the first one is a special case of the second, since for each $\epsilon>0$, we certainly can find an $a>0$ small enough such that $0<a<\min\{\mu(A),\epsilon\}$.
Here is a proof for the second question.
Let $0<\alpha<\mu(E)$. We shall find recurrence two sequences of measurable sets $(A_{n})$ and $(B_{n})$ with the following properties:

1) $A_{0}\subseteq A_{1}\subseteq\cdots\subseteq A_{n}\subseteq\cdots\subseteq B_{n}\subseteq\cdots\subseteq B_{1}\subseteq B_{0}\subseteq E$.
2) If we put 
  \begin{align*}
a_{n}&=\sup\{\mu(A): A_{n-1}\subseteq A\subseteq B_{n-1}, \mu(A)\leq\alpha\}\\
b_{n}&=\inf\{\mu(B): A_{n}\subseteq B\subseteq B_{n-1}, \mu(B)\geq\alpha\},
\end{align*}
  the sequence $(a_{n})$ is decreasing, the sequence $(b_{n})$ is increasing and we have 
  \begin{align*}
a_{n}\leq\alpha\leq b_{n},~~~~n=1,2,...
\end{align*}
3) There exist two sequences $(\epsilon_{n}),(\eta_{n})$, $\epsilon_{n},\eta_{n}\downarrow 0$ such that
  \begin{align*}
a_{n}-\epsilon_{n}<\mu(A_{n})\leq a_{n},~~~~b_{n}\leq\mu(B_{n})<b_{n}+\eta_{n},~~~~n=1,2,...
\end{align*}

Here comes the construction.
Let $(r_{n})$, $r_{n}\downarrow 0$. Put
\begin{align*}
a_{0}=\sup\{\mu(A): A\subseteq E, \mu(A)\leq\alpha\}.
\end{align*}
Then $0\leq a_{0}\leq\alpha$. Pick an $\epsilon_{0}>0$, there exists a measurable set $A_{0}$ with $A_{0}\subseteq E$ and 
\begin{align*}
a_{0}-\epsilon_{0}<\mu(A_{0})\leq a_{0}.
\end{align*}
Put 
\begin{align*}
b_{0}=\inf\{\mu(B): A_{0}\subseteq B\subseteq E, \mu(B)\geq\alpha\}.
\end{align*}
Then $\alpha\leq b_{0}\leq\mu(E)$. Pick an $\eta_{0}>0$ there exists a measurable set $B_{0}$ with $A_{0}\subseteq B_{0}\subseteq E$ and 
\begin{align*}
b_{0}\leq\mu(B_{0})<b_{0}+\eta_{0}.
\end{align*}
Put then 
\begin{align*}
a_{1}=\sup\{\mu(A): A_{0}\subseteq A\subseteq B_{0},\mu(A)\leq\alpha\}.
\end{align*}
Then $a_{0}-\epsilon_{0}<a_{1}\leq a_{0}$ and if we take $0<\epsilon_{1}\leq r_{1}$ such that $a_{0}-\epsilon_{0}<a_{1}-\epsilon_{1}$, there exists a measurable set $A_{1}$ with 
\begin{align*}
A_{0}\subseteq A_{1}\subseteq B_{0},~~~~a_{1}-\epsilon_{1}<\mu(A_{1})\leq a_{1}.
\end{align*}
Put
\begin{align*}
b_{1}=\inf\{\mu(B): A_{1}\subseteq B\subseteq B_{0},\mu(B)\geq\alpha\}.
\end{align*}
Then $b_{0}\leq b_{1}<b_{0}+\eta_{0}$ and if we take $0<\eta_{1}<r_{1}$ such that $b_{1}+\eta_{1}<b_{0}+\eta_{0}$, there exists a measurable set $B_{1}$ with 
\begin{align*}
A_{1}\subseteq B_{1}\subseteq B_{0},~~~~b_{1}\leq\mu(A_{1})<b_{1}+\eta_{1}.
\end{align*}
Now we are ready to define recursively for the rest of the sets and sequences. Suppose that we have found $A_{1},...,A_{n},B_{1},...,B_{n},\epsilon_{1},...,\epsilon_{n},\eta_{1},...,\eta_{n}$ verifying the preceding three conditions. Put
\begin{align*}
a_{n+1}=\sup\{\mu(A): A_{n}\subseteq A\subseteq B_{n},\mu(A)\leq\alpha\}.
\end{align*}
Then $a_{n}-\epsilon_{n}<a_{n+1}\leq a_{n}$ and if we take $0<\epsilon_{n+1}<r_{n+1}$ such that the inequality $a_{n}-\epsilon_{n}<a_{n+1}-\epsilon_{n+1}$ holds, there exists a measurable set $A_{n+1}$ with
\begin{align*}
A_{n}\subseteq A_{n+1}\subseteq B_{n},~~~~a_{n+1}-\epsilon_{n+1}<\mu(A_{n+1})\leq a_{n+1}.
\end{align*}
Put
\begin{align*}
b_{n+1}=\inf\{\mu(B): A_{n+1}\subseteq B\subseteq B_{n},\mu(B)\geq\alpha\}.
\end{align*}
Then $b_{n}\leq b_{n+1}<b_{n}+\eta_{n}$ and if we take $0<\eta_{n+1}<r_{n+1}$ such that $b_{n+1}+\eta_{n+1}<b_{n}+\eta_{n}$, there exists a measurable set $B_{n+1}$ with 
\begin{align*}
A_{n+1}\subseteq B_{n+1}\subseteq B_{n},~~~~b_{n+1}\leq\mu(B_{n+1})<b_{n+1}+\eta_{n+1}.
\end{align*}
Since $\epsilon_{n}\leq r_{n}$ and $\eta_{n}\leq r_{n}$, it follows that $\epsilon_{n},\eta_{n}\rightarrow 0$. These two sequences are also easily to be seen as monotone.
On the other hand, the sequences $(a_{n}),(B_{n})$ are monotone, they have limit and have
\begin{align*}
0\leq a\leq\alpha\leq b\leq\mu(E)<\infty.
\end{align*}
The sets 
\begin{align*}
A=\bigcup_{n=1}^{\infty}A_{n},~~~~B=\bigcap_{n=1}^{\infty}B_{n}
\end{align*}
are measurable and we have
\begin{align*}
A_{n}\subseteq A\subseteq B\subseteq B_{n}\subseteq E.
\end{align*}
From the third condition, we deduce that
\begin{align*}
\mu(A)=\lim_{n\rightarrow\infty}\mu(A_{n})=a\leq\alpha.
\end{align*}
Since $\mu(B_{n})<\infty$, we also deduce that 
\begin{align*}
\mu(B)=\lim_{n\rightarrow\infty}\mu(B_{n})=b\geq\alpha.
\end{align*}
Let $C$ be a measurable set such that $C\subseteq B-A$. Then $A_{n}\subseteq A\subseteq A\cup C\subseteq B\subseteq B_{n}$ for every $n=1,2,...$ If $\mu(A\cup C)\leq\alpha$, from the second and third conditions we deduce that $a_{n}-\epsilon_{n}\leq\mu(A\cup C)\leq a_{n+1}$ for all such $n$, consequently, $\mu(A\cup C)=a$. It follows that
\begin{align*}
\mu(C)=\mu(A\cup C)-\mu(A)=0.
\end{align*}
If $\mu(A\cup C)\geq\alpha$, from the second and third conditions again we deduce that $b_{n+1}\leq\mu(A\cup C)\leq b_{n}+\eta_{n}$ for every $n=1,2,...$, consequently $\mu(A\cup C)=b$, therefore $\mu(C)=\mu(A\cup C)-\mu(A)=b-a=\mu(B)-\mu(A)=\mu(B-A)$. 
Since $E$ has no atom, we deduce that $\mu(B-A)=0$, consequently $a=\mu(A)=\mu(B)=b$, therefore $\alpha=a=b$, and finally
\begin{align*}
\mu(A)=\alpha.
\end{align*}
The above proof is presented in the book by N. Dinculeanu, Vector Measures.
The author uses the following definition, which is also equivalent to the one presented in the post.

We say that a measurable set $E$ is an atom with respect to $\mu$ if $\mu(E)>0$ and if for every measurable set $A$ with $A\subseteq E$ we have either $\mu(A)=0$ or $\mu(A)=\mu(E)$. We say that $\mu$ is atomic if there exists at least one atom, and that $\mu$ is non-atomic if there exists no atom.

A: Below is a proof that follows your hint (this is also the hint given by Billingsley for the same exercise in the text Probability and Measure).
Following hint, we shall show $P(B) = a$.  By construction, it is easy to verify that $G_1, G_2, \ldots$ are disjoint and $G_n \subset A - \cup_{k < n}G_k \subset A$ for $n = 1, 2, \ldots$.  In addition, for every $n$, $\sum_{k = 1}^n P(G_k) \leq a$.
Since $P(G_k), k \geq 1$ are nonnegative, this implies $\sum_{k = 1}^\infty P(G_k) \leq a$.  It thus follows by countable additivity that
$P(\cup_k G_k) = \sum_{k \geq 1} P(G_k) \leq a$.  In the following we show that the strict inequality cannot hold, whence $\cup_k G_k$ is the
desired set $B$.
If $P(\cup_k G_k) < a$, then $P(A - \cup_k G_k) = P(A) - P(\cup_k G_k) > a - P(\cup_k G_k) > 0$, therefore by part (a) (take $\epsilon = 
a - P(\cup_k G_k) > 0$) we can choose $C \subset A - \cup_k G_k$ such that $0 < P(C) < a - P(\cup_k G_k)$.  Since $P(C) > 0$, we can choose
an $n$ sufficiently large so that $P(C) > \epsilon_n x_n$. Since $G_n \cap (A - \cup_k G_k) = G_n \cap (A \cap \cap_k G_k^c) = \varnothing$ and
$G_n \cup C \subset A - \cup_{k < n}G_k$ ($G_n \subset A - \cup_{k < n}G_k$ is straightforward, while $C \subset A - \cup_{k < n}G_k$ follows from
$\cup_{k < n}G_k \subset \cup_k G_k$ whence $C \subset A - \cup_k G_k \subset A - \cup_{k < n} G_k$). In addition, by the hypothesis,
\begin{align}
 & P(\cup_{k < n}G_k \cup (G_n \cup C)) = P(\cup_{k < n}G_k) + P(G_n) + P(C) \\
=& P(\cup_{k \leq n}G_k) + P(C) \leq P(\cup_k G_k) + P(C) < a.
\end{align}
This shows that $G_n \cup C \in \mathcal{G}_n$, where (with this notation, $x_n = \sup_{G \in \mathcal{G}_n}P(G)$)
$$\mathcal{G}_n = \{G: G \subset A - H_n, P(H_n \cup G) \leq a\}.$$
Hence $P(G_n \cup C) \leq x_n$.  Together with $P(G_n) \geq (1 - \epsilon_n)x_n$, it follows that
\begin{align*}
 & P(\cup_{k < n}G_k) + x_n \\
\leq & P(\cup_{k < n}G_k) + P(G_n) + \epsilon_n x_n \\
<& P(\cup_{k < n}G_k) + P(G_n) + P(C) \\
=& P(\cup_{k < n}G_k) + P(G_n \cup C) \leq P(\cup_{k < n} G_k) + x_n.
\end{align*}
This is a contradiction.  Therefore $P(\cup_k G_k) = a$.  This completes the proof.

Part (a) can be proved as follows (since part (b) uses part (a), we must establish an independent proof).
Let $n$ be a positive integer such that $\frac{1}{2^n}P(A) < \epsilon$.  Since $(\Omega, \mathcal{F}, P)$ is nonatomic, there exists
$B_1 \subset A$ such that $0 < P(B_1) < P(A)$, then either $0 < P(B_1) \leq \frac{1}{2}P(A)$ or $0 < P(A - B_1) \leq \frac{1}{2}P(A)$.  Let $A_1 = B_1$ if $0 < P(B_1) < \frac{1}{2}P(A)$ or $A_1 = A - B_1$ if $0 < P(A - B_1) \leq \frac{1}{2}P(A)$, then there exists
$B_2 \subset A_1$ such that $0 < P(B_2) < P(A_1)$, which either $0 < P(B_2) \leq \frac{1}{2}P(A_1)$ or $0 < P(A_1 - B_2) \leq 
\frac{1}{2}P(A_1)$. Let $A_2 = B_2$ if $0 < P(B_2) \leq \frac{1}{2}P(A_1) \leq \frac{1}{2^2}P(A)$ or $A_2 = A_1 - B_2$ if $0 < P(A - B_2)
\leq \frac{1}{2}P(A_1) \leq \frac{1}{2^2}P(A)$.  Continue in this way, we can obtain a decreasing sequence of sets $A \supset A_1 \supset A_2 \supset
\cdots$ such that $0 < P(A_k) \leq  \frac{1}{2^k}P(A), k = 1, 2, \ldots$. Take $B = A_n$, then $B \subset A$, and $0 < P(B) \leq 
\frac{1}{2^n}P(A) < \epsilon$.
