I will attempt to fill in some details for RCA 2.20.
For part c:
We wish to verify the definition $\lambda(E) = m(x + E)$, for fixed $x \in \mathbb{R}^k $ using only material in Rudin prior to this point.
First note that if $E$ is open, $x + E$ is open and thus measurable (since every Borel set is m-measurable by the Riesz Representation Theorem). Next, we know $m(E) = m(x + E)$ for open $E$ using the fact that they agree on boxes and 2.19d.
Also, we know that if $E$ is closed, then certainly $x + E$ is closed. This leads to an important observation about $F_\sigma$'s and $G_\delta$'s. Let $\{F_i\}_{i=1}^\infty$ be a collection of closed sets and let $A = \bigcup_{i=1}^\infty F_i$. We have
$$x + A = x + \bigcup_{i=1}^\infty F_i = \bigcup_{i=1}^\infty (x + F_i)$$
Since the translation of closed sets are closed, we have shown that the translation of an $F_\sigma$ is an $F_\sigma$ and thus is measurable. Similarly, the translation of a $G_\delta$ is a $G_\delta$ and thus is measurable.
That m is translation invariant for $G_\delta$'s and $F_\sigma$'s follows from the fact that m is regular and the fact that m is translation invariant on open sets, as previously discussed.
Finally, take an arbitrary measurable set $E$. By part b of the theorem there exists an $F_\sigma$, $A$, and a $G_\delta$, $B$, such that $A \subset E \subset B$ (Rudin's subset notation) and $m(B - A) = 0$. This implies that $x + A \subset x + E \subset x + B$. Since $x + A$ is an $F_\sigma$ and $x + B$ is a $G_\delta$ and $m(B-A) = m((x+B) - (x+A)) = 0$, we see immediately that part b of the theorem implies that $x + E$ is measurable and $m(E) = m(A) = m(x + A) = m(x + E)$, since m is complete.
For part e:
This proceeds more or less analogously to the previous part.
We begin with the preliminary observation that since the Borels are translation invariant (since translation is continuous) and since $T(E)$ is Borel if $E$ is Borel as noted by Rudin, the proof goes through for the restriction of $\mu$ to the Borel sigma algebra.
Next, notice that since both T and its inverse are continuous, we know that $T(E)$ is open if $E$ is open and closed if $E$ is closed.
By elementary set theory, then, $T$ maps $F_\sigma$'s to $F_\sigma$'s. Using the fact that $T$ is injective, we also see that $T$ maps $G_\delta$'s to $G_\delta$'s.
Let $E$ be an arbitrary measurable set. By part b of the theorem, there exists an $F_\sigma$, $A$, and a $G_\delta$, $B$, such that $A \subset E \subset B$ (Rudin's subset notation) and $m(B - A) = 0$. This implies $T(A) \subset T(E) \subset T(B)$. Since $T(A)$ is an $F_\sigma$ and $T(B)$ is a $G_\delta$ and $m(T(B) - T(A)) = \Delta(T)m(B-A) = 0$ (we used the fact that the proof went through for the restriction of $\mu$ to the Borel sigma algebra here), we see that part b of the theorem implies that $T(E)$ is measurable.
Suppose $\mu(S) > 0.$ Because $\mu(rE) = r^n\mu(E)$ for any measurable $E\subset \mathbb R^n$ and $r>0,$ we have $\mu(rS)\ge \mu(S)$ for $r\ge 1.$ Now $\{1\le |x|\le 2\},$ a compact subset of finite measure, contains the pairwise disjoint compact sets $S_k= (1+1/k)S, k = 1, 2, \dots $ This implies $\mu(\{1\le |x|\le 2\}) \ge \sum_k \mu(S_k) = \infty,$ contradiction.
Best Answer
Note that $$ B(0,1)\subset \overline{B}(0,1)\subset B(0,1+\delta)=(1+\delta)B(0,1) $$ for all $\delta>0$, and hence $$ m(B(0,1))\le m(\overline{B}(0,1))\le m(B(0,1+\delta))=(1+\delta)^nm(B(0,1)). $$