Show that if $\ell(I) = m^\ast(I \cap B) + m^\ast(I \cap B^c)$ holds for every $I$, then $m^\ast(A) = m^\ast(A \cap B) + m^\ast(A \cap B^c)$. 
Let $\{I_k\}_k$ be a collection of $n-$dimensional rectangles in $\Bbb R^n$. Show that if $\ell(I) = m^\ast(I \cap B) + m^\ast(I \cap B^c)$ holds for every $I$, then $$m^\ast(A) = m^\ast(A \cap B) + m^\ast(A \cap B^c) $$ holds for every $A \subset \Bbb R^n$.

Since $\ell(I) = m^\ast(I \cap B) + m^\ast(I \cap B^c)$ holds for every $I$ I have that $$m^\ast(A) \le \sum_{k=1}^\infty \ell(I_k) = \sum_{k=1}^\infty  (m^\ast(I_k \cap B) + m^\ast(I_k \cap B^c)) = \sum_{k=1}^\infty  m^\ast(I_k \cap B) + \sum_{k=1}^\infty  m^\ast(I_k \cap B^c) $$
For every $\{I_k\}_k$ that covers $A$. But now from subaddtivity I have that $$m^\ast(\bigcup_{k} I_k) \le \sum_{k=1}^\infty m^\ast(I_k)$$
so do I get that $$m^\ast(A) \le  \sum_{k=1}^\infty  m^\ast(I_k \cap B) + \sum_{k=1}^\infty  m^\ast(I_k \cap B^c) =  m^\ast(\bigcup_{k}I_k \cap B) + m^\ast(\bigcup_{k}I_k \cap B^c) $$
And since $A \subset \bigcup_{k} I_k$ I would have that $m^\ast(A) \le m^\ast(A \cap B) + m^\ast(A \cap B^c) $?
 A: Your proof requires a few changes.
For any $A\subset \mathbb{R}^d$, subadditivity yields
$$m^*(A)\leq m^(A\cap B)+m^*(A\cap B^c)$$
Thus, it suffices to show that
$$m^*(A)\geq m^*(A\cap B)+m^*(A\cap B^c)$$
If $m^*(A)=\infty$ there is nothing to do.
Suppose $\mu^*(A)<\infty$. By definition of the outer measure $m^*$,   for $m^*(A)<r$ there is a sequence of $n$-dimensional bounded rectangles $(I_k:k\in\mathbb{N})$ that cover $A$ and such that $\sum_k\mu^*(I_k)<r$.
The assumption on $B$, along with monotonicity and  subadditivity of $m^*$ gives
$$\begin{align}
r&>\sum_k m^*(I_k\cap B) +\sum_km^*(I_k\cap B^c)\geq m^*\big(B\cap \bigcup_kI_k\big)+m^*(B^c\cap\bigcup_kI_k\big)\\
&\geq m^*(B\cap A)+ m^*(B^c\cap A)
\end{align}$$
Letting $r\searrow m^*(A)$ gives the desired conclusion.
A: The inequality $m^*(A) \leqslant m^*(A \cap B) + m^*(A \cap B^c)$ is obiovus as it follows from the subadditivity of $m^*$.  It is the other inequality that requires a justification.
Moreover your proof is wrong in two places. Firstly, the equality $$\sum_{k=1}^{\infty} m^*(I_k \cap B) = m^* \left( \bigcup_{k=1}^{\infty} I_k \cap B \right)$$ does not have to be true. Secondly, the last sentence is not a correct reasoning.
