As you wrote, you have already proved $\lim_{n \rightarrow \infty} \mu^*(A_n) \leq \mu^*(A)$. It remains to show that
$\mu^*(A) \leq \lim_{n \rightarrow \infty} \mu^*(A_n)$.
Your idea to complete the proof is essentially correct, all it needs is a small adjustment.
Given any $\epsilon > 0$. We can pick a sequence of sets $\{G_i\}_{i \geq 1}$ in $\mathcal{A}$ such that $A_i\subseteq G_i$ and $\mu^*(A_i) \leq \mu(G_i) \leq \mu^*(A_i) + \epsilon/2^{i}$.
Define $H_n=\bigcap_{i=n}^\infty G_i$. Then $\{H_n\}_{n\geq 1}$ is an increasing sequence of sets in in $\mathcal{A}$.
For any $n \geq 1$, if $i \geq n$ then $A_n\subseteq A_i \subseteq G_i$. So, for any $n \geq 1$, $A_n \subseteq \bigcap_{i=n}^\infty G_i=H_n$. So we have $A_n \subseteq H_n\subseteq G_n$ and we get
$$ \mu^*(A_n) \leq \mu(H_n) \leq \mu(G_n) \leq \mu^*(A_n) + \epsilon/2^{n}$$
So we have $\lim_{n \to +\infty} \mu^*(A_n) = \lim_{n \to +\infty} \mu(H_n)$.
Since $A_n \nearrow A$, it is easy to see that that $A\subseteq \bigcup_{n\geq 1} H_n$. But $H_n \nearrow \bigcup_{n\geq 1} H_n$, so using the the continuity of measure from below, we have
$$\mu^*(A)\leq \mu\left(\bigcup_{n\geq 1} H_n\right)=\lim_{n \to +\infty} \mu(H_n)= \lim_{n \to +\infty} \mu^*(A_n)$$
We begin with a small lemma that in fact highlights part of what has already been proved in Exercise 18 item b.).
Lemma: For any $E\subset X$ there exists $B\in \mathcal{A}_{\sigma\delta}$ such that $E\subset B$ and $\mu^*(B) = \mu^*(E)$.
Proof
From Exercise 18 item a.) we know that
For any $E\subset X$ and $\epsilon > 0$ there exists $A\in \mathcal{A}_\sigma$ with $E\subset A$ and $\mu^*(E)\leq \mu^*(A) \leq \mu^*(E) + \epsilon$.
So, for each $n\in\mathbb{N}$, $n>0$, let $A_n\in \mathcal{A}_\sigma$ with $E\subset A_n$ and $\mu^*(E)\leq \mu^*(A_n) \leq \mu^*(E) + \frac{1}{n}$.
Then let $B=\bigcap_{n=1}^\infty A_n$. Then we have $B\in \mathcal{A}_{\sigma\delta}$ and $E\subset B$. Moreover, for all $n\in\mathbb{N}$, $n>0$, $B\subset A_n$ and
$$\mu^*(E)\leq \mu^*(B) \leq \mu^*(A_n) \leq \mu^*(E) + \frac{1}{n}$$
So, $\mu^*(B)=\mu^*(E)$.
Remark: $E$ don't need to be measurable. And as a consequence of item Exercise 18 item b.) $\mu^*(B\setminus E)$ may not be zero.
Exercise 19 - Let $\mu^*$ be an outer measure on $X$ induced from a finite premeasure $\mu_0$. If $E\subset X$, define the inner measure of $E$ to be $\mu_*(E) = \mu_0(X) - \mu^*(E^c)$. Then $E$ is $\mu^*$-measurable iff $\mu^*(E) = \mu_*(E)$ (Use Exercise 18).
First note that since $\mu_0$ is a finite premeasure, we have that for all $A\subset X$, $\mu^*(A)<+\infty$.
($\Rightarrow$) Suppose $E\subset X$. Note that $E^c=X\setminus E \subset X$. So,since $E$ is $\mu^*$-measurable, we have
$$\mu^*(X) = \mu^*(X\cap E) + \mu^*(X\cap E^c)=\mu^*(E) + \mu^*(E^c)$$ then, since $\mu^*(E^c)<+\infty$ and $\mu^*(X)=\mu_0(X)$, we have
$$\mu^*(E) = \mu^*(X) - \mu^*(E^c)=\mu_0(X)- \mu^*(E^c)=\mu_*(E)$$
($\Leftarrow$) If $\mu^*(E) =\mu_*(E)$ then we have, since $\mu^*(X)=\mu_0(X)$,
$$\mu^*(E) =\mu_*(E)=\mu_0(X)- \mu^*(E^c) = \mu^*(X) - \mu^*(E^c)$$
So we can conclude that
$$\mu^*(X) = \mu^*(E)+ \mu^*(E^c) \tag{1}$$
Now we apply our lemma to $E$ and $E^c$. Let $B, D \in \mathcal{A}_{\sigma\delta}$ such that $E\subset B$ and $\mu^*(B) = \mu^*(E)$ and $E^c\subset D$ and $\mu^*(D) = \mu^*(E^c)$. From $(1)$, we have
$$\mu^*(X) = \mu^*(B)+ \mu^*(D) \tag{2}$$
On the other hand, since $D \in \mathcal{A}_{\sigma\delta}$, we have that $D$ is $\mu^*$-measurable, so
$$ \mu^*(X) = \mu^*(D)+ \mu^*(D^c) \tag{3}$$
From $(2)$ and $(3)$, we get
$$ \mu^*(D)+ \mu^*(D^c)= \mu^*(B)+ \mu^*(D) $$
Since $\mu^*(D) <\infty$, we have
$$\mu^*(D^c) = \mu^*(B) \tag{4} $$
Note that since $E^c\subset D$ we have that $D^c\subset E$. So we actually have
$$D^c\subset E \subset B \tag{5}$$
and, since $D$ are $\mu^*$-measurable, $D^c$ is also $\mu^*$-measurable.
So
$$\mu^*(B)=\mu^*(D^c) + \mu^*(B\setminus D^c) \tag {6}$$
Since $\mu^*(D^c)<+\infty$ (and $\mu^*(B)<+\infty$), we have from $(4)$ and $(6)$ that
$$\mu^*(B\setminus D^c)=0$$
But from $(5)$ we have $B \setminus E\subset B\setminus D^c$, so $$\mu^*(B\setminus E)=0$$ By exercise 18 item b.), $E$ is $\mu^*$-measurable.
Best Answer
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.