Royden defines the outer measure induced by a set function $\mu : S \to [0,\infty]$ (where $S$ is an arbitrary collection of subsets of $X$) as follows: set $\mu^*(\emptyset) = 0$, and for any nonempty subset $E \subseteq X$, put
$$\mu^*(E) = \inf\left\{\sum_{k=1}^{\infty}\mu(E_k) : E \subseteq \bigcup_{k=1}^{\infty}E_k,\ \ E_k \in S\right\}$$
Let's give a name to the set in braces; I'll call it $A$. Note that $A$ is a subset of $[0, \infty]$.
In this problem, we have $S = \{\emptyset, X\}$, so in the union $\bigcup_{k=1}^{\infty}E_k$, each $E_k$ is either $\emptyset$ or $X$.
Let $E$ be any nonempty subset of $X$. (I'm assuming that $X$ itself is nonempty.)
Note that $\sum_{k=1}^{\infty}\mu(E_k)$ is either an integer or $\infty$, since the summands are all either $0$ or $1$. The only way the sum can be zero is if $E_k = \emptyset$ for all $k$, in which case $\bigcup_{k=1}^{\infty}E_k = \emptyset$, which does not contain $E$ as a subset.
So, $A$ doesn't contain $0$, hence its infimum must be at least $1$.
In fact, $A$ contains $1$, so its infimum is exactly $1$. To see this, let $E_1 = X$ and $E_k = \emptyset$ for all $k > 1$. Then $\bigcup_{k=1}^{\infty}E_k = X \cup \emptyset \cup \emptyset \cup \cdots = X$, which certainly contains $E$ as a subset. Therefore $A$ contains $\sum_{k=1}^{\infty}\mu(E_k) = \mu(X) + \mu(\emptyset) + \mu(\emptyset) + \cdots = 1 + 0 + 0 + \cdots = 1$.
We conclude that $\mu^*(E) = 1$ for any nonempty subset $E \subseteq X$, and of course $\mu^*(\emptyset) = 0$.
For the next step we want to know which $E \subseteq X$ satisfy the Caratheodory criterion
$$\mu^*(B) = \mu^*(B \cap E) + \mu^*(B \cap E^c)$$
for all $B \subseteq X$.
In particular, setting $B = X$, we see that such a set $E$ must satisfy
$$1 = \mu^*(X) = \mu^*(E) + \mu^*(E^c)$$
The terms on the RHS are either $0$ or $1$, so exactly one of them must be $0$ and the other must be $1$. In particular, since one of the terms must be zero, this means that either $E$ or $E^c$ must be $\emptyset$, and consequently either $E^c$ or $E$ must be $X$.
The above reasoning shows that the only subsets of $X$ which satisfy the Caratheodory criterion with respect to $\mu^*$ are $\emptyset$ and $X$, i.e. $S = \{\emptyset, X\}$ is the $\sigma$-algebra of measurable sets with respect to the measure induced by $\mu^*$.
You have taken the $F_n$ as measurable sets with $E_n\subseteq F_n$ such that $\mu(E_n)=\mu(F_n)$ through regularity. Moreover without loss of generality you can assume that the $F_n$ are an increasing sequence. Indeed, you can increase $F_n$ by redefining it as its union with $F_{n-1}$. The regularity property holds for this union still since
$$ \mu(F_n\cup F_{n-1}\setminus E_n)\leq \mu(F_n\setminus E_n)+ \mu(F_{n-1}\setminus E_n)\leq \mu(F_n\setminus E_n)+\mu(F_{n-1}\setminus E_{n-1})=0.$$
Then you have
$$ \bigcup_{n=1}^\infty E_n \subseteq \bigcup_{n=1}^\infty F_n,$$
so you can use property 2 of your outer measure to get
$$ \mu\left(\bigcup_{n=1}^\infty E_n\right) \leq \mu\left(\bigcup_{n=1}^\infty F_n\right)=\lim_{n\to\infty}\mu(F_n)=\lim_{n\to\infty}\mu(E_n),$$
with the equalities following from your observation that the claim holds for measurable sets and the regularity condition. To get the lower inequality, note that for each $n$ you trivially have $E_n\subseteq \bigcup_{j=1}^\infty E_j$ and so by property 2 of your outer measure
$$\mu(E_n)\leq \mu\left(\bigcup_{j=1}^\infty E_j\right),$$
taking the limit $n\to\infty$ then gives the result.
Best Answer
Proof: Since $E \in M$ and $E \cap F=\emptyset$, we have \begin{align} \mu^*(A \cap (E \cup F)) &= \mu^*(A \cap (E \cup F) \cap E) + \mu^*(A \cap (E \cup F) \cap E^c) =\\ &= \mu^*(A \cap E) + \mu^*(A \cap F) \end{align} $\square$
Now for the main result:
Since for all $n$, $A \cap E_n \subseteq \lim_k A \cap E_k$, we have $\lim_n \mu^*( A \cap E_n) \leqslant \mu^*(\lim_n A \cap E_n)$.
Now, let $E= \lim_n E_n = \bigcup_n E_n$. Let us define $F_0 = E_0$ and, for each $n$, $F_{n+1} =E_{n+1} \setminus \bigcup_{k=0}^n E_k$. Then the sets $F_n$ are disjoint set and for all $n$, $F_n \in M$. Moreover $E_n= \bigcup_{k=0}^n F_k$ and $E= \lim_n E_n = \bigcup_n E_n= \bigcup_n F_n$.
So we have $$\lim_n \mu^*( A \cap E_n) \leqslant \mu^*(\lim_n A \cap E_n)= \mu^* \left (\bigcup_n ( A \cap F_n) \right ) \leqslant \sum_n\mu^*(A\cap F_n)$$
where the last step is $\sigma$-sub-additivity of $\mu^*$. So we have:
$$\lim_n \mu^*( A \cap E_n) \leqslant \mu^*(\lim_n A \cap E_n)\leqslant \lim_k\sum_{n=0}^k\mu^*(A\cap F_n) \tag{1}$$
Using the Lemma and induction, we have that for all $k$, $$\sum_{n=0}^k\mu^*(A\cap F_n)= \mu^*\left (A \cap \bigcup_{n=0}^k F_n \right ) = \mu^*(A \cap E_k) \tag{2}$$
Combining $(1)$ and $(2)$:
$$\lim_n \mu^*( A \cap E_n) \leqslant \mu^*(\lim_n A \cap E_n)\leqslant \lim_k\sum_{n=0}^k\mu^*(A\cap F_n) = \lim_k\mu^*(A \cap E_k) $$
So $\lim_n \mu^*( A \cap E_n) = \mu^*(\lim_n A \cap E_n)$