About the proof: $\mathbb Q$ is not locally compact $$\mathbb Q \text{ is not locally compact.}$$
Proof by Brian M. Scott on this answer:

Suppose that $C$ is a compact subset of $\Bbb Q$ with $0$ that contains an open nbhd of $0$; then there is an $\epsilon>0$ such that $(-\epsilon,\epsilon)\cap\Bbb Q\subseteq C$. Let $f:\Bbb Q\to\Bbb R$ be the identity map; $f$ is continuous, so $f[C]$ is compact, and $f[C]\supseteq f[(-\epsilon,\epsilon)\cap\Bbb Q]=(-\epsilon,\epsilon)\cap\Bbb Q$. But $\Bbb R$ is Hausdorff, so the compact set $f[C]$ must be closed, and therefore $f[C]\supseteq\operatorname{cl}_{\Bbb R}\big((-\epsilon,\epsilon)\cap\Bbb Q\big)=[\epsilon,\epsilon]$. This, however, is impossible, since $f[C]$ is countable, and $[\epsilon,\epsilon]$ is not.



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*Why there exists and $\epsilon>0$ such that $(-\epsilon,\epsilon)\cap\Bbb Q\subseteq C$ ?


and 


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*Why $f[C]\supseteq[\epsilon,\epsilon]$ with $f[C]$ is countable, and $[\epsilon,\epsilon]$ is not
is the contradiction?
please someone enlighten me! 
Thanks in advance


For the first bullet, that is the definition of the statement "$C$ contains an open nbhd of $0$". 
For the second bullet, you need some set theory: every subset of a countable set is countable; and every image of a countable set under a function is countable. Since $\mathbb{Q}$ is countable, it's subset $C$ is countable, and the image $f(C)$ is countable, so the subset $[-\epsilon,\epsilon] \subset f(C)$ is countable. However, $[-\epsilon,\epsilon]$ is not countable, which gives the contradiction.
