strange metric $d(x,y) = ||x|| + ||y||$ if $x\ne y$, $d(x,y) = 0$ if $x = y$. Let $d : \mathbb{R}^n \times \mathbb{R}^n \to [0, \infty]$ be defined by
$$
 d(x,y) = \left\{ \begin{array}{ll} 0 & : ~ x = y \\ ||x|| + ||y|| & : ~ x \ne y \end{array} \right.
$$
where $||\cdot ||$ denotes the usual norm of $\mathbb{R}^n$.


*

*Show that $d$ is a metric.

*Draw the $\varepsilon$-Spheres $B_{\varepsilon}(x_0) := \{ x \in \mathbb{R}^2 ~|~ d(x,x_0) < \varepsilon \}$ for $x_0 = (0,0)$ and $x_0 = (1,1)$ and $\varepsilon = \frac{1}{2}, 1, \frac{3}{2}$.

*Characterize the open, closed and compact sets with respect to this metric.

*Is $(\mathbb{R}^n, d)$ complete?
Number 1) is simple, for 2) I got:
If $x_0 = (0,0)$ then
$$
d(x,x_0)= \left\{ \begin{array}{ll} 
0 & \textrm{ for } x = (0,0) \\
\sqrt{x^2 + y^2} & \textrm{ otherwise } \end{array} \right. 
$$
and if $x_0 = (1,1)$ then
$$
  d(x,x_0) =  \left\{ \begin{array}{ll} 
0 & \textrm{ for } x = (1,1) \\
\sqrt{2} + \sqrt{x^2 + y^2} & \textrm{ otherwise } \end{array} \right. 
$$
and the pictures are simple spheres with the point $x_0$ in the sphere ($x_0 = (0,0)$) or isolated outside ($x_0 = (1,1)$).
But with 3) I have my problems, i conjecture that 
$$
 B_{\varepsilon}(x) \quad \textrm{ is open iff } \quad ||x|| - \varepsilon > 0
$$
and going on I know that finite intersections of open sets are open, but then had I got all open sets by this construction? And what about the other properties, how can I characterize them, do you have any hints?
 A: For every $x \ne 0$ we can find $0 < \epsilon < \|x\|$ so that $B_\epsilon(x) = \{x\}$. Therefore, for every $x \ne 0$, $\{x\}$ is open. Also, evidently $B^d_\epsilon (0) = B_\epsilon(0)$; i.e. all Euclidean balls around $0$ are open in the topology induced by $d$ (denote it with $\tau_d$ from now on).
Thus, every open in $\tau_d$ is of the form:
$$B_\epsilon(0) \cup S$$
with $\epsilon \ge 0, B_\epsilon(0)$ an Euclidean ball around $0$ (remark the empty case $\epsilon = 0$) and $S \subseteq X \setminus \{0\}$. 
The closed sets are then characterised as $\complement(B_\epsilon(0)) \cap \complement(S)$ with $B_\epsilon(0)$ and $S$ as above. Now $\complement (S)$ contains $0$ but is otherwise an arbitrary subset of $X$. Since $\complement(B_\epsilon(0))$ does not contain $0$ if $\epsilon > 0$, the only requirement remaining for $C \subseteq S$ to be closed in this case, is that it be contained in the complement of some $B_\epsilon(0)$. If $\epsilon = 0$, then since $\complement(B_\epsilon(0)) = X$, the only remaining requirement is that $0 \in C$.
This is summarized by saying that $C \subseteq X$ is closed iff:
$$0 \in C \quad \text{or}\quad \inf \{\|c\|: c \in C\} =: \operatorname{dist}(C, 0) > 0$$
The compact sets are easily shown to be all finite $S \subseteq X$.
Finally $(\Bbb R^n, d)$ is complete since a Cauchy sequence $(x_n)_n$ either takes finitely many values (and trivially converges) or has for $\epsilon >0$ that $d(x_n,x_m) < \epsilon \implies d(x_n,0) <\epsilon$ (since $x_n \ne x_m$ infinitely often), and we conclude $\displaystyle \lim_{n \to \infty} x_n = 0$ in this case.
EDIT: Grateful thanks to fgp for pointing out clumsy errors in my analysis of the problem.
A: Regarding (2), the $B_e(x_0)$ are not all spheres, I believe - some of them contain just a single point. You have $d(x,y) \geq ||x||$, so $B_\epsilon(x_0)$ is $\{x_0\}$ if $||x_0|| \geq \epsilon$, since you then have $d(x_0,y) \geq \epsilon$ for all $y \neq x_0$. Thus $$
  B_\epsilon(x_0) = \{x_0\} \bigcup \{y : ||y|| < \epsilon - ||x_0||\}
$$
The second set is empty if $\epsilon \leq ||x_0||$.
Regarding (3) - You need to characterize the open sets of the topology induced by $d$. Thus, all $B_\epsilon(x)$ are open by definition. And so are all (finite or infinite) unions of sets $B_{\epsilon_i}(x_i)$. I guess what they want you to do is to find a more direct way to characterize the open sets than that. Usually, the set of all open sets is not easily described without resorting to at least infinit unions, but here you can do much better. Look at the balls $B_{||x||}(x)$ for arbitrary $x$, in particular...
One interpretation of this metric, btw, is that to go from $a$ to $b$, you always have to go from $a$ to $0$ first, and then from $0$ to $b$. Thus, for every point $a$, the closest "neighbour" of $a$ that isn't $a$ itself is always $0$, and its distance is $||a||$. The matric is therefore sometimes called french railroad metric, alluding to the (surely exaggerated!) claim that to get from one city to another by railroad in france, you always have to go via paris.
