Let $\, X := \mathbb{R}^3\Big/_{\sim} \,$ where $\, \sim \,$ is defined as: $\,x \sim y \iff x = y \quad \lor \quad \lVert x\rVert = \lVert y \rVert > 4$. Say wheter the canonical map $\, \pi :\mathbb{R}^3 \rightarrow X$ is open or not and if $X$ is a second-countable space or not.
If $\pi$ is open then we can easily say that $X$ is a second-countable space and to me it seems, intuitively, that $\pi$ should be open, but i can't make any progress towards building a formal proof of this (neither find a counter-example). Any hints would be appreciated.
Hint
$\pi$ is open if and only if for any open subset $U \subset \mathbb R^3$ $\pi^{-1}(\pi(U))$ is open.
Now you can prove (by considering the cases $\Vert x \Vert <, =, > 4$) that for an open ball $B(x,r)$ centered on $x$ of radius $r < \Vert x \Vert$, $\pi^{-1}(\pi(B(x,r)))$ is open. Which proves that $\pi$ is open.
The "interesting case" is when $B(x,r)$ intersects the sphere centered on the origin of radius $4$. In particular if $\Vert x \Vert = 4$, you can prove that $$\pi^{-1}(\pi(B(x,r))) = B(x,r) \cup \left(B(0,4+r) \setminus \overline{B}(0,4)\right)$$ which is the union of two open subsets.