Prove that $C=\bigcup_{n=1}^{\infty}I_n$ is a compact subset of a space with a rail metric 
Let $I_n$ - line segment connecting the points $(0,0)$ and $\Big( \frac{\cos (\frac{2\pi}{n})}{n}, \frac{\sin (\frac{2\pi}{n})}{n} \Big)$ for $n\in \mathbb N \setminus \{0\}$. Prove that $C=\bigcup_{n=1}^{\infty}I_n$ is a compact subset of a space with a rail metric.

I think $C$ is a compact in rail metric because we can cover each of the line segments with a railway ball starting at $(0,0)$ and radius $1$. There will be countabless infinite numbers of these balls, because $|C|=|\mathbb N|=\aleph_0$. That's why $C$ is compact.
However I think that my proof is not very formal and results more from graphic intuition. Anyone would like to say how to prove it "elegantly"?
 A: Note that the sequence of end points converges to $(0,0)$ and in any space, a convergent sequence with its limit together form a compact set, as any neighbourhood of the limit already contains all but finitely many terms of the sequence by convergence, and this keeps on being true with the extra line segments because with the endpoint a neighbourhood will also contain the line segment. And the finitely many segments not in the neighbourhood of the origin form a finite union of compact segments so can also by covered by finitely many members of the cover.
Note that your proof is totally wrong: you have to show that every open cover of the set has a finite subcover. Showing that one finite cover exists proves nothing at all.
A: The rail metric on $\Bbb{R}^2$ (defined as discussed in the comments above and not as in the Wikipedia article on metric spaces) agrees with the Euclidean metric on any line through the origin, so (as compactness is a topological property), each $I_n$ is compact. Also, it is easy to check from the definitions, that, if $B_R(\mathbf{v}, r)$ is the open ball around $\mathbf{v}$ of radius $r$ under the rail metric and $\mathbf{0} \in B_R(\mathbf{v}, r)$, then $r > \| v\|$ and $B_r(\mathbf{v}, r) \supseteq B_E(\mathbf{0}, r - \|v\|)$ (the open ball of radius $r - \|v\|$ around the origin $\mathbf{0}$ under the Euclidean metric). So, if $\bigcup_{n=1}^\infty I_n$ is covered by a set $\mathcal{C}$ of open balls in the rail metric, then some $B_0 \in \mathcal{C}$ has $\mathbf{0} \in B_0$ and hence contains $B_E(\mathbf{0}, \delta)$ for some $\delta > 0$. But $B_E(\mathbf{0}, \delta)$ contains all but finitely many of the $I_n$. As each of the finitely many $I_n$ that are not contained in $B_E(\mathbf{0}, \delta)$ is compact, their union is compact and hence is covered by some finite subset $\{B_1, \ldots, B_k\}$ of $\mathcal{C}$. So $\{B_0, \ldots B_n\}$ is a finite subset of $\mathcal{C}$ that covers $\bigcup_{n=1}^\infty, I_n$. This gives us that $\bigcup_{n=1}^\infty I_n$ is compact in the rail metric.
Aside: under the post office metric, any singleton other than $\{\mathbf{0}\}$ is open and the $I_n$ and $\bigcup_{n=1}^\infty I_n$ are not compact. However, the union of the endpoints of the $I_n$ is compact under that metric (by a similar argument to the above).
