Railway metric, help showing that the space is bounded but not compact I am studying metric spaces and I came across this exercice that I was able to solve only partially. Any help and hints will be appreciated. 
Let $X= \{ x \in \mathbb{R}^2 | \mid \mid x \mid \mid \leq 1 \}$ on which we consider the railway metric defined by
$ d(x,y)= \mid \mid x-y \mid \mid$ if $x$ and $y$ are on the same line through the origin and 
$d(x,y)= \mid \mid x \mid \mid + \mid \mid y \mid \mid$ otherwise.
I could show that this is indeed a metric, I could show that the space is complete but now I have to show that $(X,d)$ is bounded but not compact. Also, is this space separable ? Since $X$ is homeomorphic to $\mathbb{R}^2$, I would say yes, but I am not sure.
Any hints or ideas on how to show that this space is bounded and not compact ?
Thank you a lot for your help
 A: The distance between any two points is never greater than $2$. And the sequence$$\left(\cos\left(\frac1n\right),\sin\left(\frac1n\right)\right)_{n\in\mathbb N}$$has no convergent subsequence. This follows from the fact that the distance between any two points of the sequence is equal to $2$. So, no subsequence is a Cauchy sequence. In particular, no subsequence converges.
A: Let $0$ be the origin in $\Bbb R^2.$ For each line $L$  in $\Bbb R^2$ thru $0,$ the set $L'= (L\cap X)$ \ $\{0\}$ is open in $(X,d)$ and is non-empty.
And if $L_1\ne L_2$ then $L_1'$ and $L'_2$ are disjoint, because  two unequal lines  thru $0$ in $\Bbb R^2$ meet $only$  at $0.$ A dense subset in $(X,d)$ would have at least one point in each $L',$ and the set of all $L'$ is uncountable.
So $(X,d)$ is not  separable .
A compact metric space $(S,d)$ is separable. One way to prove this is:  For  each $n\in \Bbb N$ the set of all open balls of radius $1/n$ is an open cover of $S,$ so let $S_n$ be a finite subset of $S$ such that $F_n=\{B_d(x,1/n):x\in S_n\}$ covers $S.$ Then $\cup_{n\in \Bbb N}S_n $is  countable and dense. 
