Let $O_n$ be the sequence such that $E \subset O_n$ and $m^*(O_n-E) \to 0$. (I am going to use $m^*$ for the outer measure).
By subadditivity, we have that
$$m^*(A) \le m^*(A \cap E) + m^*(A \cap E^c).$$
For the other inequality, since open sets are Lebesgue measurable (by either definition - you should check this), we have that for any $A \subset \mathbb{R}^d$, we have that
$$m^*(A) = m^*(A \cap O_n) + m^*(A \cap O_n^c).$$
We note that $E \subset O_n$, so $m^*(A \cap O_n) \ge m^*(A \cap E)$ and
$$m^*(E^c \cap A) \le m^*(O_n^c \cap A) + m^*((O_n - E) \cap A) \Rightarrow m^*(O_n^c \cap A) \ge m^*(E^c \cap A) - m^*((O_n - E) \cap A).$$
Thus, we get that
$$m^*(A) \ge m^*(A \cap E) + m^*(A \cap E^c).$$
For the reverse direction, assume that $m^*(A) = m^*(A \cap E) + m^*(A \cap E^c)$. By the definition of the outer measure, we have that
$$m^*(E) = \inf \left\{ \sum m^*(R_i) : E \subset \bigcup R_i, R_i \text{ is a rectangle}\right\}.$$
Let $(\{R^n_i\})_n$ be a sequences of covers of $E$ such that
$$m^*\left(\bigcup_i R_i^n\right) \le \sum_i m^*(R^n_i) \le m^*(E) + \frac{1}{n}.$$
Using this define $O_n = \bigcup_i R_i^n$ and apply the Caratheodory condition with $O_n$ to get
$$m^*(E) + \frac{1}{n} \ge m^*(O_n) = m^*(O_n \cap E) + m^*(O_n \cap E^c) = m^*(E) + m^*(O_n \cap E^c). $$
This implies that
$$ \frac{1}{n} \ge m^*(O_n \cap E^c) = m^*(O_n - E). $$
Thus, we have the needed sequence of sets. Thus, we are done.
Best Answer
One important key is that $m^*(\mathcal O - A) = m^*(\mathcal O \cap A^c)$.
Two the Lebesgue Measure is defined via volume of open boxes, and open sets, like $\mathcal O$, are made up of open boxes.
Three in (2) $m^*(A)\leq m^*(A \cap X) + m^*(A \cap X^c)$ is always true. So the only relevant proof is one for $m^*(A) \geq m^*(A \cap X) + m^*(A \cap X^c)$.
Using these three it should be simple to get an idea, the rest is technical.
Note: a box is always open in the context below.
Start by proving (2) for every $A$ is a box, and $X$ only range every open box of $\mathbb R^d$. It is simple as then $A \cap X$ is another box, and $A \cap X^c$ is made up of finite decomposition of boxes $I_n$, which should give you a hint about what the Lebesgue measures of those two parts are. So you now have a finite sequence of boxes $I_n$ and a single box $I$ such that $m^*(A \cap X) = m^*(I)$ and $m^*(A \cap X^c) = \Sigma m^*(I_n)$, These follows from definition.
So $m^*(A \cap X)+ m^*(A \cap X^c) = \Sigma m^*(I_n)+ m^*(I)$.
But $I_n \cup I = A$ you use also covers $A$ (remember $A$ is a box) perfectly, meaning $m^*(A) = \Sigma m^*(I_n)+ m^*(I)$.
Then we are going to extend it to every $X \subseteq \mathbb R^d$, instead of just boxes. $A$ is still just a box, however. But that is trivial, because we can cover $X$ with sequences of open boxes, which will give an upper bound of the measure $m^*(A \cap X)$ and $m^*(A \cap X^c$. By the result we just proved, and sigma subadditivity, we can obtain
$$m^*(A \cap X)+ m^*(A \cap X^c) \leq m^*(A).$$
Any open set is a countable union of open intervals. It then extends that (2) holds for ANY open sets. (Measurable sets form a $\sigma$-algebra for any measure. The proof is application of subadditivity).
Now (1) $\implies$ (2) is trivial. (Note "boxes" are open in our context) As we know (2) holds for $\mathcal O$. $\forall \varepsilon > 0$, we can find $\mathcal O_\varepsilon$
$$ m^*(O_\varepsilon) - m^*(A)$$ $$ = m^*((A) \cup (\mathcal O_\varepsilon-A)) - m^*(A)$$ $$ \leq m^*(A)+m^*(\mathcal O_\varepsilon -A) - m^*(A)$$ $$ = m^*(O_\varepsilon-A) <\varepsilon$$.
Then with $A \cap X \subseteq \mathcal O \cap X$, we get $m^*(A \cap X) \leq m^*(\mathcal O_\varepsilon \cap X)$, same for $X^c$, it follows that
$$m^*(A \cap X)+ m^*(A \cap X^c)$$ $$ \leq m^*(\mathcal O_\varepsilon \cap X)+ m^*(\mathcal O_\varepsilon \cap X^c) $$ $$ = m^*(\mathcal O_\varepsilon) = m^*(A) + \varepsilon \text{ ((2) for all open sets)}$$
$\varepsilon$ is arbitrary, meaning $m^*(A \cap X)+ m^*(A \cap X^c) \leq m^*(A)$, completing (1) $\implies$ (2).
For the other direction, you will need to create $\mathcal O$ from open boxes, specifically those used in the definition of $m^*(A)$. Since $m^*(A) = \inf\{$ total measure of sequence of boxes$\}$, meaning for every $\varepsilon$, we can find a sequence of boxes $\mathcal O = \bigcup I_n$ which measure $m^*(\mathcal O) - m^*(A) < \varepsilon$.
Sorry for a super long answer, but the three key above, and subaddivity of Lebesgue Measure. Hope it helps.