Q:

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.