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.
Your proof for the first part is correct. For the second part, you basically showed that if $x_n \to a$, then $f(x_n)\to L$ for some $L$. The problem though is that, a priori, this $L$ that arose depends on the sequence you picked. You have to show there aren't two sequences $x_n \to a$ and $y_n \to a$, with $f(x_n)\to L_1$ and $f(y_n) \to L_2$ and $L_1 \neq L_2$.
I'll sketch the idea assuming we are in $\mathbb{R}$. WLOG, suppose $L_2>L_1$. Then I claim we can choose $\epsilon>0$ so that for any $x\in B(L_2;\epsilon)=(L_2-\epsilon,L_2+\epsilon)$ and for any $y\in B(L_1;\epsilon)=(L_1-\epsilon,L_1+\epsilon)$, we have $|x-y|>\epsilon$. This follows easily since $|x-y|\geq x-y>(L_2-\epsilon)-(L_1+\epsilon)=(L_2-L_1)-2\epsilon$, which is greater than $\epsilon$ if $\epsilon<\frac{L_2-L_1}{3}$.
Choose such an $\epsilon$. By hypothesis, there exists $\delta_1>0$ such that if $x,y \in B(a;\delta_1)\backslash\{a\}$, then $|f(x)-f(y)|<\epsilon$. Since $x_n \to a$ and $f(x_n)\to L_1$, there exists $N_x$ such that for all $n\geq N_x$, we have $x_n \in B(a;\delta)\backslash\{a\}$ and $f(x_n)\in B(L_1;\epsilon)$. Similarly, there exists $N_y$ such that for all $n\geq N_y$, we have $y_n \in B(a;\delta)\backslash\{a\}$ and $f(y_n) \in B(L_2;\epsilon)$. Take any $n_0 \geq \max\{N_x,N_y\}$. Then $x_{n_0},y_{n_0}\in B(a;\delta_1)\backslash\{a\}$, but
$$
|f(x_{n_0})-f(y_{n_0})|\geq f(x_{n_0})-f(y_{n_0})>(L_2-L_1)-2\epsilon>\epsilon,
$$
a contradiction.
In the higher dimension case, it's an easy argument that $\epsilon>0$ can be chosen small so that $d(B(L_1;\epsilon),\, B(L_2;\epsilon))>\epsilon$.
Best Answer
Assume that $\mu^*(A)+\mu^*(A^c)\le 1$, let $\varepsilon > 0$ and set $\delta := \tfrac\varepsilon 6$. Then we find $E_i,F_j\in\mathcal A$ such that $A\subset\bigcup_i E_i$, $A^c\subset\bigcup_j F_j$, and $\mu^*(A)>\sum_i|E_i|-\delta$ as well as $\mu^*(A^c)>\sum_j|F_j|-\delta$. Hence, $$ 1\,\ge\,\mu^*(A)+\mu^*(A^c)\,>\,\sum_i|E_i| + \sum_j|F_j| - 2\delta. $$ Let $n\in\Bbb N$ be so large such that $\sum_{i=n}^\infty|E_i| < \delta$ and $\sum_{j=n}^\infty|F_j| < \delta$. Now, put $E:=\bigcup_{i<n}E_i$ and $F:=\bigcup_{j<n}F_j$. Then, since $A\backslash E\subset\bigcup_{i=n}^\infty E_i$, we have $\mu^*(A\backslash E)\le\sum_{i=n}^\infty|E_i| < \delta$. Similarly, $\mu^*(A^c\backslash F) < \delta$. Now, $$ E\cup F = (E\backslash F)\cup (F\backslash E)\cup(E\cap F). $$ Since $E,F\in\mathcal A$, we also have $E\backslash F,F\backslash E,E\cap F\in\mathcal A$, and so \begin{align*} \mu^*(E\cup F) &= \mu^*(E\backslash F) + \mu^*(F\backslash E) + \mu^*(E\cap F)\\ &= \mu^*(E)-\mu^*(E\cap F) + \mu^*(F)-\mu^*(E\cap F) + \mu^*(E\cap F)\\ &= \mu^*(E)+\mu^*(F)-\mu^*(E\cap F). \end{align*} Also, $X\backslash(E\cup F)\subset(A\backslash E)\cup (A^c\backslash F)$, which implies $$ \mu^*(X\backslash(E\cup F))\le\mu^*(A\backslash E)+\mu^*(A^c\backslash F) < 2\delta. $$ Thus, we obtain $$ 1 = \mu^*(E\cup F) + \mu^*(X\backslash(E\cup F)) < \mu^*(E)+\mu^*(F) - \mu^*(E\cap F) + 2\delta. $$ Finally, this and $A\triangle E = (A\backslash E)\cup(E\cap A^c)\subset (A\backslash E)\cup(A^c\backslash F)\cup(E\cap F)$ yield $$ \mu^*(A\triangle E)\le 2\delta+\mu^*(E\cap F) < 4\delta + \mu^*(E)+\mu^*(F) - 1\,\le\,4\delta + \sum_i|E_i| + \sum_j|F_j| - 1 < \varepsilon. $$