Your proof for $\fbox{$\Rightarrow$}$ is correct.
Now let us look at the $\fbox{$\Leftarrow$}$ part.
First, note that this part is true only if $c- b < +\infty$, this means, if $c$ and $b$ are finite. In fact, if $b= -\infty$ or $c = +\infty$, the equation $|A|+ |(b, c)\setminus A|= c -b$ becomes trivially true for all subsets $A \subseteq (b,c)$.
Now, assume $c- b < +\infty$, $A \subseteq (b,c)$ and $|A|+ |A^c|= c - b$ (where $A^c$ means $(b,c) \setminus A$). Let us prove $A$ is Lebesgue measurable.
Note that, since $|(b,c)| < +\infty$, we have that $|A| <+\infty$ and $|A^c| <+\infty$.
Given any set $S \subseteq (b,c)$, such that $|S| <+\infty$, then there is a Borel set $T \subseteq (b,c)$, such that $S \subseteq T$ and $|S| =|T|$. This is a direct consequence of the definition of outer measure and its is true for all $S$ such that $|S| +\infty$, no matter $S$ is Lebesgue measurable or not. (Attention, in general, $|S| =|T|$ does not imply $|T \setminus S| = 0$).
The simplest way to prove the $\fbox{$\Leftarrow$}$ part is apply the above property to $A$ and $A^c$ and use the fact that $|A|+ |A^c|= c − b$.
Here is a detailed proof:
Since $|A| <+\infty$, let $E \subseteq (b,c)$ be a Borel set, such that $A \subseteq E$ and $|A| =|E|$. Since $E$ is a Borel set, we have
$c -b = |(b,c)| = |E| + |E^c| $.
Since $|A|+ |A^c|= c - b$, we have
$$|E^c| = (c-b) -|E| = (|A|+ |A^c|) - |A| = |A^c|$$
In a similar way, since $|A^c| <+\infty$, let $F \subseteq (b,c)$ be a Borel set, such that $A^c \subseteq F$ and $|A^c| =|F|$. Since $F$ is a Borel set, we have
$c -b = |(b,c)| = |F| + |F^c| $.
Since $|A|+ |A^c|= c - b$, we have
$$|F^c| = (c-b) -|F| = (|A|+ |A^c|) - |A^c| = |A|$$
Now, note that $F^c$ and $E$ are Borel sets and that $F^c \subseteq A \subseteq E$. So
$$|E| = |F^c| + |E\setminus F^c| $$
But $|E|=|A|=|F^c|$. So
$$|E\setminus F^c| =0$$
Since $A \subseteq E$, we have that $(A \setminus F^c) \subseteq (E\setminus F^c)$, and so $|A \setminus F^c| =0$.
So we have that $F^c$ is a Borel set, $F^c \subseteq A$ and $|A \setminus F^c| =0$. So $A$ is Lebesgue measurable.
Best Answer
Take $E\subseteq \Bbb{R}$ a bounded measurable set that is not Borel.
Then by regularity of the Lebesgue measure there is a bounded $F_{\sigma}$ set $F$ and a bounded $G_{\delta}$ set $G$ such that $F\subseteq E \subseteq G$
and $F=\bigcup_nF_n$ and $G=\bigcap_n G_n$ and $G_1$ has finite measure.
Take $O_n=\bigcap_{k=1}^nG_n$ and $K_n=\bigcup_{k=1}^nF_n$
and $\mu(O_n \setminus K_n) \to \mu(G\setminus F)=0$