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.
Best Answer
$\newcommand{\esssup}{\mathrm{ess\,sup}}$$\newcommand{\essrng}{\mathrm{ess\,range}}$ You want to prove that $\esssup(f) \geq \sup (\essrng(f))$. For this, it's enough to show that if $w \in \essrng(f)$, then $\esssup(f) \geq |w|$.
Hint: You can do this by contradiction: Assume there is $w_0 \in \essrng(f)$ such that $\esssup(f)<|w_0|$ and try to find a contradiction by playing around with the definitions of $\esssup$ and $\essrng$.
Solution: (Try not to look at this until you've tried to solve it by yourself)