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
Yes, (ii) is definitely true, because of the scaling property of Lebesgue measure. However, proving (ii) directly isn't easy because we're dealing with an infinite measure. The typical way of dealing with this is to exploit $\sigma$-finiteness of Lebesgue measure.
What I would suggest is to define for each $k\in\Bbb{Z}$, $E_k:=[k,k+1)$, and let $\mathcal{L}_k$ the set of all Borel measurable $A\subset E_k=[k,k+1)$ such that $\lambda(cA)=|c|\lambda(A)$. It is immediate that the intervals $\mathcal{I}=\{[a,b)\,:\, k\leq a<b<k+1\}$ is a $\pi$-system on $E_k$ and it is contained in $\mathcal{L}_k$. Points (i) and (iii) hold for very trivial reasons. For point (ii), let $A\in\mathcal{L}_k$ be arbitrary. Then, \begin{align} \lambda(c\cdot (E_k\setminus A))&=\lambda(cE_k)-\lambda(cA)\\ &=|c|\lambda(E_k)-|c|\lambda(A)\\ &=|c|\lambda(E_k\setminus A) \end{align} The subtraction is valid due to the sets having finite measure. Hence, $E_k\setminus A\in\mathcal{L}_k$, and thus $\mathcal{L}_k$ is a $\lambda$-system on $E_k$. By Dynkin's theorem, $\mathcal{L}_k$ equals all Borel subsets of $E_k=[k,k+1)$.
Now, we don't have to worry about proving $\mathcal{L}$ is a $\lambda$-system. We can directly show that it equals all Borel sets because for any Borel set $A\subset\Bbb{R}$, we have \begin{align} \lambda(c\cdot A)&=\lambda\left(c\cdot \bigcup_{k\in \Bbb{Z}}E_k\cap A\right)\\ &=\lambda\left(\bigcup_{k\in\Bbb{Z}}c\cdot (E_k\cap A)\right)\\ &=\sum_{k\in\Bbb{Z}}\lambda\left(c\cdot (E_k\cap A)\right)\\ &=\sum_{k\in\Bbb{Z}}|c|\lambda(E_k\cap A)\tag{$*$}\\ &=|c|\lambda(A) \end{align} where we have repeatedly used countable additivity of measures. Also, $(*)$ is because $E_k\cap A$ is a Borel subset of $E_k$, and hence by what we showed above, the formula holds for it. This proves the formula holds for all Borel subsets. (And from here, since the Lebesgue $\sigma$-algebra is the completion, we can deduce the formula holds for all Lebesgue measurable sets as well).
Note by the way that this reduction to the finite measure case is also one of the possible proofs for the "uniqueness" of Lebesgue measure (i.e it is the unique translation-invariant measure defined on Lebesgue $\sigma$-algebra, finite on compact sets, and normalized to 1, i.e measure of $[0,1)$ is $1$).
Also, essentially the proof outline works for the more general situation of proving that in $\Bbb{R}^n$, for any linear $T:\Bbb{R}^n\to\Bbb{R}^n$ and Lebesgue measurable set $A\subset\Bbb{R}^n$, we have $\lambda(T(A))=|\det T|\lambda(A)$.