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 is in the right direction. Here I copied it and I changed some points in it to make it a complete proof.
Theorem 1.9 - Suppose that $(X,M,\mu)$ is a measure space. Let $\mathcal{N} = \{N\in M:\mu(N) = 0\}$ and $\overline{M} = \{E\cup F: E\in M, F\subset N, N\in\mathcal{N}\}$. Then $\overline{M}$ is a $\sigma$-algebra and there is a unique extension $\overline{\mu}$ of $\mu$ to a complete measure on $\overline{M}$.
Claim 1 - $\overline{M}$ is a $\sigma$-algebra
proof:
i.) Since $M$ is a $\sigma$-algebra, $\emptyset\in M\subset \overline{M}$ so $\emptyset\in \overline{M}$.
ii.) Suppose $B\in\overline{M}$ then there is an $E\in M$ such that $B = E\cup F$ where $F\subset N$ and $N\in\mathcal{N}$. Then
\begin{align*}
X\setminus B &= X\cap B^c\\
&= X\cap (E\cup F)^c\\
&= X\cap ((E^c\cap N^c\cap F^c)\cup(F^c\setminus N^c))\\
&= X\cap ((E^c\cap N^c)\cup (F^c\cap N))\\
&= X\cap ((E\cup N)^c\cup (N\setminus F))
\end{align*}
Since $(E\cup N)^c\in M$ and $(N\setminus F)\subset N\in M$ with $\mu(N) = 0$, then $B^c\in\overline{M}$.
iii.) Let $\{B_j\}_{1}^{\infty}\in\overline{M}$ then for each $j$ there is an $E\in M$ such that $B_j = E_j\cup F_j$ where $F_j\subset N_j$ and $\mu(N_j) = 0$. So, $$\bigcup_{1}^{\infty}B_j = \bigcup_{1}^{\infty}(E_j\cup F_j) = \bigcup_{1}^{\infty}E_j \cup \bigcup_{1}^{\infty}F_j$$ Note that $\bigcup_{1}^{\infty}F_j\subset \bigcup_{1}^{\infty}N_j$ and $\mu\left(\bigcup_{1}^{\infty}N_j\right) = 0$. So we have $\bigcup_{1}^{\infty}B_j\in\overline{M}$. Therefore $\overline{M}$ is a $\sigma$-algebra.
Claim 2 - There is a unique extention $\overline{\mu}$ of $\mu$ to a complete measure on $\overline{M}$
Proof:
We first need to show that $\overline{\mu}$ is well-defined. Suppose $E\cup F\in\overline{M}$, set $\overline{\mu}(E\cup F) = \mu(E)$. This is well-defined since if $E_1\cup F_1 = E_2\cup F_2$ where $F_j\subset N_j\in\mathcal{N}$. Then we know that $E_1\subset E_1\cup F_1 = E_2\cup F_2 = E_2\cup N_2$ then $E_1\subset E_2\cup N_2$ and so by monotonicity $\mu(E_1)\leq \mu(E_2) + \mu(N_2) = \mu(E_2)$. Also, since $E_2\subset E_2\cup F_2 = E_1\cup F_1 = E_1\cup N_1$ then again by monotonicity $\mu(E_2)\leq \mu(E_1)+\mu(N_1) = \mu(E_1)$. So we have $\mu(E_1) = \mu(E_2)$. Thus $\overline{\mu}$ is well-defined.
I now need to show that $\overline{\mu}$ is a complete measure on $\overline{M}$ and that $\overline{\mu}$ is the only measure on $\overline{M}$ that extends $\mu$.
Step 1 - Show $\overline{\mu}$ is a measure.
Proof:
i.) $\overline{\mu}(\emptyset) = \overline{\mu}(\emptyset \cup \emptyset)= \mu(\emptyset) = 0$ (here we took $E=\emptyset$ and $F= N=\emptyset$).
ii.) Let $\{A_n\}_{1}^{\infty}\in\overline{M}$, disjoint, then there is an $\{E_n\}_{1}^{\infty}\in M$ and a sequence $\{F_n\}_{1}^{\infty}\subset N\in\mathcal{N}$ such that $A_n = E_n\cup F_n$ for all $n$. Note that $\{E_n\}_{1}^{\infty}$ is a family of disjoint sets in $M$. Note also from part (iii.) of claim 1, $\cup_{1}^{\infty}F_n\subset \cup_{1}^{\infty}N_n$ and $\mu(\cup_{1}^{\infty}N_n)=0$. Thus,
\begin{align*}
\overline{\mu}\left(\bigcup_{1}^{\infty}A_n\right) = \overline{\mu}\left(\bigcup_{1}^{\infty}E_n\cup F_n\right) &=
\overline{\mu}\left(\bigcup_{1}^{\infty}E_n\cup \bigcup_{1}^{\infty}F_n\right)\\
&= \mu\left(\bigcup_{1}^{\infty}E_n\right)=\\
&= \sum_{1}^{\infty}\mu(E_n)=\\
&= \sum_{1}^{\infty}\overline{\mu}(E_n\cup F_n) \\
&= \sum_{1}^{\infty}\overline{\mu}(A_n)
\end{align*}
Thus $\overline{\mu}$ is a measure.
Step 2 - Show $\overline{\mu}$ is a complete measure.
Proof:
Let $A\subset X$ and suppose there is an $B\in\overline{M}$ such that $A\subset B$ and $\overline{\mu}(B) = 0$. Since $B\in\overline{M}$, we know we can write
$B = E\cup F$ where $E\in M$ and $F\subset N\in M$ with $\mu(N) = 0$. Since $\overline{\mu}(B) = 0$ and $\overline{\mu}(B) = \mu(E)$, we have $\mu(E)=0$. So
$E\cup N \in M$ and $\mu(E\cup N)=0$.
Now note that $A=\emptyset \cup A$, $\emptyset \in M$ and $A\subseteq B = E\cup F \subseteq E\cup N$ and $E\cup N \in M$ and $\mu(E\cup N)=0$. So $A \in \overline{M}$ and $\overline{\mu}(A)=\mu(\emptyset)=0$. So $\overline{\mu}$ is complete.
Step 3: Show $\overline{\mu}$ is a unique extension of $\mu$
3.1. $\overline{\mu}$ is an extension of $\mu$.
Given $A \in M$, we have that $A=A\cup\emptyset$ so $A=A\cup F$ where $F\subset N$, $N\in M$ and $\mu(N)=0$ (just take $F=N=\emptyset$). Thus, $A \in \overline{M}$ and
$\overline{\mu}(A) =\overline{\mu}(A\cup \emptyset)=\mu(A)$.
So we proved that $M \subseteq \overline{M}$ and $\overline{\mu}$ is an extension of $\mu$.
3.2. Uniqueness.
Suppose $\nu$ is an extension of $\mu$ to $\overline{M}$. Let $A\in\overline{M}$, then there is $E\in M$, $N\in M$ with $\mu(N)=0$ and $F\subseteq N$ such that $A=E\cup F$. Then, since $E\in M$ and $E\cup N\in M$,
$$\mu(E)=\nu(E)\leqslant \nu(A)=\nu(E\cup F)\leqslant \nu(E\cup N)=\mu(E\cup N)
\leqslant \mu(E)+\mu(N)=\mu(E)$$
So $\nu(A)=\mu(E)=\overline{\mu}(A)$. So $\nu = \overline{\mu}$.
Best Answer
Your proof is essentially correct for the first part. I will reword it following Follands and adding some commments and including the other parts.
Proof a.) implies b.) If $E\subset\mathbb{R}$ and $E\in M_{\mu}$ then, by theorem 1.18, for each $n\in \mathbb{N}$, $n>0$, there is an open set $U_n$ such that $E\subset U_n$ and $$\mu(E)+ \frac{1}{n} \geq \mu(U_n) \geq \mu(E)$$ Since $E\subset \bigcap_{1}^{\infty} U_n$, we have, for each $n\in \mathbb{N}$, $n>0$, $$\mu(E)+ \frac{1}{n} > \mu(U_n)\geq \mu\left(\bigcap_{1}^{\infty} U_n\right) \geq \mu(E)$$ Thus, $$\mu(E)= \mu\left(\bigcap_{1}^{\infty} U_n\right)$$
Set $V = \bigcap_{1}^{\infty}U_n$. $V$ is a $G_{\delta}$, $E \subset V$ and $\mu(V)=\mu(E)$. Set $N_1 = V \setminus E$ then, since $E,V \in M_\mu$, we have $N_1 \in M_\mu$. Since $E\subset V$, we have $E=V\setminus N_1$.
If $\mu(E)<+\infty$ then $\mu(N_1)=\mu(V)-\mu(E) =0$
Remark: Folland's book stops here and leaves the general case to the reader (exercise 25). However, since the countable union of $G_\delta$ may not be a $G_\delta$, it is not straight forward to use the fact $\mu$ is $\sigma$-finite to extend the result for sets of finite measures to the general case. In fact, proving the general case is more like re-doing the proof completely than simply extending the result for sets of finite measures.
General Case:
If $E\subset\mathbb{R}$ and $E\in M_{\mu}$, then since $\mu$ is $\sigma$-finite, we have that there are $\{E_k\}_1^\infty \subset M_\mu$ such that $\{E_k\}_1^\infty$ is a family of disjoint sets, $\mu(E_k) <+\infty$, for all $k$ and $E=\bigcup_1^\infty E_k$. Then, by theorem 1.18, we have, for each $k$, for each $n\in \mathbb{N}$, $n>0$, there is an open set $U_{k,n}$ such that $E_k\subset U_{k,n}$ and $$\mu(E_k)+ \frac{1}{n}\frac{1}{2^k} \geq \mu(U_{k,n}) \geq \mu(E_k) \tag{1}$$ Let $U_n=\bigcup_1^\infty U_{k,n}$ then $U_n$ is open, $E\subset U_n$ and from $(1)$ we have $$\mu(E)+ \frac{1}{n} =\sum_1^\infty \mu(E_k)+ \sum_{k=1}^\infty\frac{1}{n}\frac{1}{2^k} \geq \sum_{k=1}^\infty\mu(U_{k,n})\geq \mu(U_n) \geq \mu(E) \tag{2}$$ Note also that, using that $\mu(E_k) <+\infty$, for each $k$, we have $$\mu(U_n \setminus E)\leq \sum_{k=1}^\infty\mu(U_{k,n}\setminus E) \leq \sum_{k=1}^\infty\mu(U_{k,n}\setminus E_k)= \sum_{k=1}^\infty \left (\mu(U_{k,n})-\mu( E_k) \right)\leq \sum_{k=1}^\infty\frac{1}{n}\frac{1}{2^k} =\frac{1}{n} \tag{3} $$ Set $V = \bigcap_{1}^{\infty}U_n$. $V$ is a $G_{\delta}$ and $E \subset V$. From $(2)$, we have, for each $n\in \mathbb{N}$, $n>0$,
$$\mu(E)+ \frac{1}{n} \geq \mu(U_n)\geq \mu(V) \geq \mu(E)$$ So $\mu(V)=\mu(E)$. Since $E,V \in M_\mu$, we have $V \setminus E \in M_\mu$ and, from $(3)$, we have, for each $n\in \mathbb{N}$, $n>0$, $$\mu(V\setminus E) \leq \mu(U_n \setminus E)\leq \frac{1}{n} $$ So $\mu(V\setminus E)=0$. Set $N_1 = V \setminus E$ then, since $E\subset V$, we have $E=V\setminus N_1$ (where $V$ is a $G_\delta$ and $\mu(N_1)=0$).
Thus a.) implies b.).
Proof a.) implies c.) It can be proved in a way similar to the way we proved that a.) implies b.), by using the second part of theorem 1.18.
However there is another elegant way to prove it:
If $E\in M_{\mu}$, then $E^c \in M_{\mu}$, so by the previous item [a.) implies b.)] we know that $E^c=V\setminus N_1$, where $V$ is a $G_\delta$ and $\mu(N_1)=0$.
So, since $E^c=V\setminus N_1=V\cap N_1^c$, then, by taking complement, we have $E=V^c\cup N_1$.
Now, note that the complement of a $G_\delta$ is an $F_\sigma$, so $V^c$ is an $F_\sigma$. Just take $H=V^c$ and $N_2=N_1$.
Proof b.) implies a.)
Suppose $E\subset\mathbb{R}$ and $E=V\setminus N_1$, where $V$ is a $G_\delta$ and $\mu(N_1)=0$. Since $V$ is a $G_\delta$, we have that $V \in M_\mu$ and, since $\mu$ is complete, we also have $N_1\in M_\mu$. So $$E=V\setminus N_1\in M_n$$
Proof c.) implies a.)
Suppose $E\subset\mathbb{R}$ and $E=H \cup N_2$, where $H$ is a $F_\sigma$ and $\mu(N_2)=0$. Since $F$ is a $F_\sigma$, we have that $F \in M_\mu$ and, since $\mu$ is complete, we also have $N_2\in M_\mu$. So $$E=H \cup N_2\in M_n$$