Suppose first that $E$ is of finite measure. Since $E$ is measurable, for each $\varepsilon > 0$ there exists by definition $E \subseteq G$ open with $|G \setminus E| < \varepsilon$. Similarly we have $F \subseteq E$ open with $|E \setminus F| < \varepsilon$ (just observe that $E^\complement$ is measurable and use the former). Thus, for each $n \geq 1$ we have $F_n \subseteq E \subseteq G_n$ with $F_n$ closed, $G_n$ open and $|E\setminus F_n|,|G_n \setminus E| < \frac{1}{n}$. Now set $F = \bigcup_n F_n$ and $G = \bigcap_n G_n$ which are $F_\sigma$ and $G_\delta$ respectively (in particular, Borel sets), and note that
$$
|E \setminus F| \leq |E \setminus F_n| < \frac{1}{n} \to 0
$$
and
$$
|G \setminus E| \leq |G_n \setminus E| < \frac{1}{n} \to 0.
$$
So we have $|G \setminus F| = |G| - |F| = 0$, because $|G| = |E| = |F|$.
Finally, suppose that $E$ has infinite measure an let $E_n = E \cap \overline{B}(0,n)$. By the previous result, we have Borel sets $F_n \subseteq E_n \subseteq G_n$ with $|F_n| = |E| = |G_n|$ and so if $F = \bigcup_nF_n$, $G = \bigcup_n G_n$, we get $F \subseteq E \subseteq G$. To conclude, recall that since $F$ and $G$ are increasing union of measurable sets,
$$
|F| = \lim_n |F_n| = \lim_n|E_n| = |E| = +\infty
$$
and similarly for $G$.
The first property is called 'outer regularity' and the second 'inner regularity'. First of all, you can not take $A_\delta$ as defined above, as you have already noted. Also the proof of the second equality (i.e. your argument in the section third inequality) is false, because it can happen that $K_n$ is not a subset of $A$! (Take your example: Then $A_\delta = \mathbb{R}$ and $K_n$ will approximate the measure of $\mathbb{R}$.)
To prove the 'outer regularity' use the Caratheodory construction of the Lebesgue measure: If $\lambda(A) < \infty$, then we can find $(a_n,b_n] = \prod_{i=1}^d (a_{n,i},b_{n,i}] $ disjoint ($d$-dimensional) cubes with
$$\sum_{k=1}^\infty \lambda ((a_n,b_n]) < \lambda(A) + \varepsilon/2$$
Thus, we can take $U= \bigcup_{n=1}^\infty (a_n,b_n+ t_n \varepsilon)$. (Note $ A \subset \bigcup_{n=1}^\infty (a_n,b_n] \subset U$ and if $t_n$ is taken appropriate, e.g. $t_n = 2^{-n-2d-1} \max\{1,b_{n,i} -a_{n,i}\}^{-(d-1)} $ we get $\lambda(U) < \lambda(A) + \varepsilon$.)
The proof ot the 'inner regularity' is more complicated. First recall that any set $\mathcal{M}^*$ is the completion of the Borel-$\sigma$-algebra. Thus any $A \in \mathcal{M}^*$ can be written as $A= B \cup M$ with a Borel set $B$ and $M \subset N$ with a Borel-nullset $N$. Thus, we only have to prove the 'inner regularity' for Borel-sets. Define the system
$$\mathcal{D} := \{ B \in \mathcal{B}(\mathbb{R}^d) : B \text{ is inner regular}\}.$$
- Your argument shows that open sets are in $\mathcal{D}$.
- Check that $\mathcal{D}$ is a Dynkin-system.
- Since the set of all open sets is a $\cap$-stable generator of the Borel-$\sigma$-algebra we can conclude that $\mathcal{D} = \mathcal{B}(\mathbb{R}^d)$.
I additionally added a prove of 2: Note that $\mathbb{R}^d$ is an open set. If $(A_n)_{\in \mathbb{N}} \subset \mathcal{D}$ are disjoint, then we can take compact $K_n \subset A_n$ with $\lambda(A_n) < \lambda(K_n) + \varepsilon 2^{-n}$. Since a finite union of compact sets is compact, but in general not an infinite union, we have to truncate the sums as follows.
First case: Now if $\lambda(\cup_{n=1}^\infty A_n) = \sum_{n=1}^\infty \lambda(A_n) = \infty$, then for any $K>0$ there exists $N$ such that $$\sum_{n=1}^N \lambda(A_n) >K.$$ Thus $\sum_{n=1}^N \lambda(K_n) > K-\varepsilon$. So taking the compact set $K = \cup_{k=1}^N K_n$ shows that the supremum is $>K-1$. Since $K>0$ was arbitrary, we get that the supremum is $\infty$.
In the second case, we have for some $N \in \mathbb{N}$, because the series is convergent, that $$\lambda(\cup_{n=1}^\infty A_n) < \sum_{n=1}^N \lambda(A_n) + \varepsilon.$$
Thus
$$\lambda(\cup_{n=1}^\infty A_n) < 2 \varepsilon + \lambda(\cup_{n=1}^N K_n) $$
and therefore we can take the compact set $K = \cup_{n=1}^N K_n$. This proves that $\cup_{n=1}^\infty A_n \in \mathcal{D}$.
Let $A,B \in \mathcal{D}$ with $B \subset A$. For the next argument we need that both $B$ and $A$ have finite measure. Of course, this is satisfied if the sets are bounded. Thus take $$A_i = A \cap \prod_{i=1}^d (n_i,n_i+1] \quad \text{and} \quad B_i = B \cap \prod_{i=1}^d (n_i,n_i+1]$$ with $n_i \in \mathbb{Z}$ instead of $A$ and $B$ to get bounded sets. (If $A_i \setminus B_i \in \mathcal{D}$. then also the union of this disjoint sets is in $\mathcal{D}$ by the previous argument. Moreover, show that also $A_i,B_i \in \mathcal{D}$.)
So we can assume that both $B$ and $A$ are bounded. Now take an open set $U$ with $B \subset U$ and $\lambda(U \setminus B) < \varepsilon$ (this is possible, because we already know that $\lambda$ is outer regular) and a compact set with $K \subset A$ and $\lambda(A \setminus K) < \varepsilon$. Define $L = K \setminus U$. Then $L$ is compact and
$$\lambda( (A \setminus B) \setminus L) \le \lambda(U \setminus B) + \lambda(A \setminus K) < 2 \varepsilon.$$
Therefore $A \setminus B \in \mathcal{D}$.
Best Answer
Any set of diameter $d$ is contained in a closed ball of diameter $2d$. So a cube of diameter $d$ is covered by a ball of diameter $2d$. The volume of this ball is proportional to the volume of the cube since both volumes scale as $d^n$, so this proves the claim.