$\lim_{k \to \infty}\Vert x_{k} \Vert = +\infty$ then $\lim_{k \to \infty}\Vert f(x_{k}) \Vert = +\infty$ if $f$ is an homeomorphism? 
Problem. Prove or disprove:
(a) If $f: \mathbb{R}^{n} \to \mathbb{R}^{n}$ is a homeomorphism and $(x_{k})_{k \in \mathbb{N}}$ is a sequence such that
  $$\lim_{k \to \infty}\Vert x_{k} \Vert = +\infty,\;\mathrm{then}\;\lim_{k \to \infty}\Vert f(x_{k}) \Vert = +\infty.$$
(b) If $X \subset \mathbb{R}^{n}$ is not compact, then there is $f: X \to \mathbb{R}$ continuous and unbounded.
(c) Let $A \subset \mathbb{R}^{m}$, $f:A \to \mathbb{R}^{n}$ and $a \in \mathrm{Int}(A)$. Suppose that the directional derivative exist for every $u \in \mathbb{R}^{m}\setminus \{0\}$ and define
  $$T(u) = \begin{cases}0,&\quad \mathrm{if}\;u=0 \\ D_{u}(a),&\quad \mathrm{if}\;u\neq 0. \end{cases}$$
  Then $T$ is a linear transformation.

For (b): True. Take $X = \mathbb{R}^{n}$ and $\pi$ the projection over the first coordinate.
For (c): I have not thought of a counterexample yet, but for $T$ to be linear, $f$ must be differentiable, right?
I'm in doubt about item (a). It seems to be true, but I couldn't prove it. Probably, I'm not seeing any little detail. Can someone help me?
 A: (a) is true
If the image under $f$ of a sequence $(x_k)$ diverging to $\infty$ is not diverging to $\infty$, the image of a subsequence $(x_{k_n})$ under $f$ is bounded. By extracting again a subsequence, you can suppose that $(v_n)$ is a diverging subsequence whose image $(y_n)=(f(v_n))$ converges to $y$. As $f$ is supposed to be an homeomorphism, $f^{-1}$ exists and is continuous. In contradiction with the fact that $(y_n)$ converges to $y$ while $(f^{-1}(y_n))$ diverges to $\infty$.
(b) is true
If $X$ is not compact, it is either not closed or unbounded.
If $X$ is not closed, it exists $a \in \overline X\setminus X$. Then $x \mapsto \dfrac{1}{\Vert x -a\Vert}$ is continuous on $X$ and unbounded.
And if $X$ is supposed to be unbounded the map $x \mapsto \Vert x \Vert$ is continuous on $X$ and unbounded.
(c) is wrong
The map 
$$h(x,y)=\begin{cases}\frac{x^2 y}{x^6+y^2} & \text{ if } (x,y) \ne (0,0)\\ 
0 &  \text{ if }(x,y) = (0,0)\end{cases}$$
has directional derivatives in all directions at the origin that are not all vanishing. However both partial derivatives at the origin vanish. Hence the associated map $T$ can’t be linear (if it was it would be always vanishing).
