Let $(B_n)_{n \in \mathbb{N}}$ be a sequence of pairwise disjoint $m^*$-measurable sets. Is a subsequence $C_k \subset B_k$ also $m^*$measurable? I am new to Measure theory and the title is a question I came up with that I would intuitively answer as yes, having only a background in real and multivariable Analysis. To refine my question:

Let $\Omega$ be a set and $m^*$ be an outer measure on $\Omega$.  A subset $B \subset \Omega$ is $m^*$-measurable $: \iff$ $m^*(A)= m^*(A \cap B) + m^*(A \cap B^C)$ for all $A \subset \Omega$. Define:$$ \mathcal{M}:= \lbrace B \subset \Omega: B \text{ is $m^*$-measurable} \rbrace  $$

A very relevant theorem I was exposed to in class is that $\mathcal{M}$ is indeed a $\sigma$-Algebra, but more than that $m^*$ restricted to $\mathcal{M}$ is a measure.
That is, if I take $(B_n)_{n \in \mathbb{N}} \subset \mathcal{M}$ a sequence of pairwise disjoint $m^*$ measurable sets, then I will have that $$ m^* \left( \bigcup_{k \in \mathbb{N}} B_k \right) = \sum_{k \in \mathbb{N}} m^*(B_k) $$
I wonder now if I produce any subsequence of $(B_n)_{n \in \mathbb{N}}$ say $C_k \subset B_k$ for every $k \in \mathbb{N}$ will the above equality also hold for $(C_k)_{k \in \mathbb{N}}$?
If my understanding of the theorem is good, then it is enough to show that $C_k \in \mathcal{M}$ i.e. $$ m^*(A)=m^*(A \cap C_k) + m^*(A \cap C_k^c) ,  \ \forall A \subset \Omega $$
But I would not know how to tackle this given that $$ m^*(A)= m^*(A \cap B_k) + m^*(A \cap B_k^c), \ \forall A \subset \Omega $$
 A: 
I wonder now if I produce any subsequence of $(B_n)_{n \in \mathbb{N}}$ say $C_k \subset B_k$ for every $k \in \mathbb{N}$ will the above equality also hold for $(C_k)_{k \in \mathbb{N}}$?

$C_k\subset B_k$does not mean that $C_k$ will be $\mathcal{M}$ measurable, so your approach will not work. You need to use facts about the outer measure.
Let $m$ denote the outer measure restricted to the sigma algebra $\mathcal{M}$ so that $m$ is a measure. Then to prove this result we will use that $$m^*(E)=\inf\left\{\sum_{i=0}^\infty m(F_i)\  :\ E\subset\bigcup_{i=1}^\infty F_i\text{ and } F_i \in \mathcal{M} \right\}.$$
Suppose I had just two sets $C_1\subset B_1$, $C_2\subset B_2$. Then we need only show that $$m*(C_1\cup C_2)\geq m^*(C_1)+m^*(C_2),$$ since the outer measure is countably subadditive. Let $\epsilon>0$, suppose that $\{G_j\}_{j=0}^\infty$ is a sequence of $m$-measurable sets approximating $C_1\cup C_2$, and $C_1\cup C_2\subset\bigcup_{j=0}^\infty G_j$
$$m^*(C_1\cup C_2)+\epsilon \geq \sum_{j=0}^\infty m(G_j).$$ Then $G_j\cap B_1 \subset B_1$ is a sequence of measurable sets containing $C_1$, and $G_j\cap B_2 \subset B_2$ is a sequence of measurable sets containing $C_2$. By the additivity of the measure and the fact that the $B_j$ are disjoint, $m(G_j)\geq m(G_j\cap B_1)+m(G_j\cap B_2)$. But then, by definition of $m^*$ we have that $$m^*(C_1\cup C_2)+\epsilon \geq m^*(C_1)+m^*(C_2).$$ Since $\epsilon$ was arbitrary, this proves the inequality, and hence $$m^*(C_1\cup C_2)=m^*(C_1)+m^*(C_2).$$ This proof can be generalized to work for any countable union of sets.
