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.
Since this is old but not properly answered, I'll knock it out. (See also the comment by Callus, which I didn't notice until I was done writing this)
Now
$$L(c_{k+1}\overline{u}_{k+1}+\cdots+c_{n}\overline{u}_{n}) = \overline{0}$$
which means that $c_{k+1}\overline{u}_{k+1}+\cdots+c_{n}\overline{u}_{n}\in\operatorname{Ker}(L)$. This is the part that I'm getting stuck at and I don't know how I should go on with proving the linear independence.
Since $\overline{u}_1,\overline{u}_2,\dots,\overline{u}_k$ are a basis for the kernel, there are constants $c_1,c_2,\dots,c_k$ so that
$$c_{k+1}\overline{u}_{k+1}+\cdots+c_{n}\overline{u}_{n} = -c_1\overline{u}_1-\cdots-c_k\overline{u}_k$$
$$c_1\overline{u}_1+\cdots+c_k\overline{u}_k+c_{k+1}\overline{u}_{k+1}+\cdots+c_{n}\overline{u}_{n} = 0$$
Now we use the linear independence of the $u_i$ to conclude that all of the $c_i$ are zero. That includes $c_{k+1}$ through $c_n$, so $L(\overline{u}_{k+1}),\dots,L(\overline{u}_{n})$ are linearly independent.
The proof that $L(\overline{u}_{k+1}),\dots,L(\overline{u}_{n})$ span the image was good all along, and this completes the the proof of the theorem.
Best Answer
Let $l<k$ be the dimension of $Y=T(\mathbb R^k).$ Then there exists a bijective linear map $S:\mathbb R^l\to Y.$ If $K\subset Y$ is compact, then $S^{-1}(K)$ is a compact subset of $\mathbb R^l.$ It thus suffices to show $m_k(S(E))=0$ for every compact subset $E$ of $\mathbb R^l.$
Let such an $E$ be given. Choose $a>0$ such that $E\subset [-a,a]^l.$ Let $N$ be a large positive integer. Subdivide $[-a,a]$ into $2N$ subintervals $I_j$, each of length $a/N.$ Then
$$\tag 1 [-a,a]^l = \bigcup I_{j_1}\times I_{j_2} \cdots \times I_{j_l},$$
where $j_1, \dots ,j_{l}$ each run independently through $1,\dots, 2N.$ There are exactly $(2N)^l$ cubes on the right of $(1).$
Now every linear map on $\mathbb R^l$ is Lipschitz, so $S$ is Lipschitz. Thus there exists $M>0$ such that $|S(y)-S(x)|\le M|y-x|$ for all $x,y \in \mathbb R^l.$
Let $C$ be any of the cubes in the union in $(1).$ Then $\text {diam }C=\sqrt la/N.$ The Lipschitz condition then implies $\text {diam }S(C) \le M\sqrt l a/N.$ Take any point $p\in S(C).$ Then $S(C)$ is contained in a regular $k$-cube centered at $p,$ of side length $2M\sqrt l a/N.$ The volume of this last cube equals $(2M\sqrt l a/N)^k.$
We're in good shape: $S(E)$ is contained in the union of $(2N)^l$ $k$-cubes whose $k$-measures are each $(2M\sqrt l a/N)^k.$ The volume of this union is thus bounded above by
$$(2N)^l \frac{(2M\sqrt l a)^k}{N^k} = (2M\sqrt l a)^kN^{l-k}.$$
This is true for every $N.$ Because $l<k,$ the right side $\to 0$ as $N\to\infty.$ This shows the $k$-volume of $S(E)$ is $0,$ and we're done.