Finding a set with probability satisfying an inequality, part 2 - slightly changed Extended from Finding a set with probability satisfying an inequality, with additional assumptions added.
I am requesting a hint, not necessarily a complete solution. The intuition behind this solution is more important than the solution itself.
Let's say I have a sample space $\Omega$ with $\mathbb{P}(\Omega) = 1$, $\mathbb{P}(\{x\}) = 0$ for every singleton, and for any $D \subset \Omega$ with $\mathbb{P}(D) > 0$, there is a set $E \subset D$ with $0 < \mathbb{P}(E) < \mathbb{P}(D)$. **
I have two disjoint sets $A, B \subset \Omega$ with $0 < \mathbb{P}(A) < \dfrac{1}{3}$ and $\mathbb{P}(B) > \dfrac{2}{3}- \mathbb{P}(A)$. We also know that $$\mathbb{P}(A) = \sup\left\{ \mathbb{P}(A^{\prime}):A^{\prime} \subset \Omega, \mathbb{P}(A^{\prime}) < \dfrac{1}{3}\right\}\text{.}$$
Is there a way I can construct a set $C \subset \Omega$, based on $A$, $B$, and $\Omega$ (using set operations), so that $\mathbb{P}(C) \in \left[\dfrac{1}{3}, \dfrac{2}{3} \right]$?
I tried working with $F:= A \cup (\Omega \setminus (A \cup B))$, and showed that $\mathbb{P}(A) \leq \mathbb{P}(F) < \dfrac{2}{3}$, but that's not sufficient.
** I believe it can be shown that the conditions "$\mathbb{P}(\{x\}) = 0$," and "for any $D \subset \Omega$ with $\mathbb{P}(D) > 0$, there is a set $E \subset D$ with $0 < \mathbb{P}(E) < \mathbb{P}(D)$" are equivalent.
 A: Given the specified conditions, there will always exist an event $C$ such that
$
\left\{{\large{\frac{1}{3}}}\le P(C)\le {\large{\frac{2}{3}}}\right\}
$.

Proof:

A preliminary lemma . . .

Lemma:

For all $\epsilon > 0$ and all $Y$ with $P(Y) > 0$ there exists 
$X\subset Y$ such that $0 < P(X) < \epsilon$.

Proof of the lemma:

Let $\epsilon > 0$ and let $Y$ be an event with $P(Y) > 0$.

Let $n$ be a positive integer such that ${\large{\frac{1}{n}}} < \epsilon$.

Applying the hypothesis, we can find a partition of $Y$ into $n$ sets $Y_1,...,Y_n$ such that $P(Y_k) > 0$ for all $k\in\{1,...,n\}$.

But then from
$$
P(Y_1)+\cdots +P(Y_n)=P(Y) \le 1
$$
it follows that $P(Y_k) \le {\large{\frac{1}{n}}}$ for some $k\in\{1,...,n\}$.

This completes the proof of the lemma.

Returning to the main proof . . .

Let $S=\left\{X{\,\colon\,}P(X)\ge{\large{\frac{2}{3}}}\right\}$, and let $w=\inf\{P(X){\,\colon\,}X\in S\}$.

Necessarily $w\ge{\large{\frac{2}{3}}}$.

Choose $X_1,X_2,X_3,...\in S$ such that ${\displaystyle{\lim_{n\to\infty}}} P(X_n)=w$.

For each positive integer $n$, let $Y_n={\displaystyle{\bigcap_{k=1}^n}} X_k$.

First suppose $Y_n\not\in S$ for some positive integer $n$.

Let $m$ be the least positive integer such that $Y_m\not\in S$.

Note that $Y_1\in S$, hence $m > 1$.

From $Y_m=Y_{m-1}\cap X_m$ we get
$$
P(Y_m)
=
P(Y_{m-1}\cap X_m)
=
P(Y_{m-1})+P(X_m)-P(Y_{m-1}\cup X_m)
\ge
\frac{2}{3}+\frac{2}{3}-1
=
\frac{1}{3}
$$
and from $Y_m\not\in S$ we have $P(Y_m) < {\large{\frac{2}{3}}}$, hence we can let $C=Y_m$ and we're done.

Next suppose $Y_n\in S$ for all positive integers $n$.

Let $Y={\displaystyle{\bigcap_{n=1}^\infty}} X_n$.

For all $n$ we have $Y_n\subseteq X_n$, so $P(X_n)\ge P(Y_n)\ge w$.

Then ${\displaystyle{\lim_{n\to\infty}}} P(Y_n)=w$, hence since
$$
Y_1\supseteq Y_2\supseteq Y_3\supseteq\cdots
$$
we get $P(Y)=w$.

Suppose $w > {\large{\frac{2}{3}}}$.

Applying the lemma, there exists $Z\subset Y$ such that $0 < P(Z) \le w-{\large{\frac{2}{3}}}$.

Then
$$
P(Y{\setminus}Z)=P(Y)-P(Z)=w-P(Z) > w-\left(w-{\small{\frac{2}{3}}}\right)={\small{\frac{2}{3}}}
$$
so $Y{\setminus}Z\in S$, contradiction, since
$P(Y{\setminus}Z)=P(Y)-P(Z) < P(Y)=w$.

It follows that $w = {\large{\frac{2}{3}}}$, hence since $P(Y)=w$, we can let $C=Y$ and we're done.

This completes the proof.

Note:

As regards your statement suggesting that the conditions

*
(1)$\;\;P(\{x\}) = 0$ for all $x\in\Omega$.$\\[4pt]$

(2)$\;\;$For all $D \subset \Omega$ with $P(D) > 0$, there is a set $E \subset D$ with $0 < P(E) < P(D)$.
are equivalent, the example I used in my answer to your prior question

*
Finding a set with probability satisfying an inequality
shows otherwise, since for that example, condition $(1)$ holds, but condition $(2)$ fails.
