Show that this quotient space is not Hausdorff

Question

I have been at this problem for hours with a friend, drawing pictures and discussing and giving ourselves massive headaches. At this point we suspect that the key is that an open set containing a y-axis point must also contain x-axis points, but we haven't been able to actually find two concrete points that cannot be separated via open sets (and, there is a good chance that we are just completely wrong about the x-axis/y-axis thing).

I need some kind of assistance, please and thank you.

Answer 1

Hint: Try taking two points on the $y$-axis. To understand what open sets around them will look like, consider the following. If $V\subseteq Y$ is an open neighborhood of $(0,y)$ for some $y$, then $V=f(U)$ where $U=f^{-1}(V)\subseteq\mathbb{R}^2$ is open and $(0,y)\in U$. In particular $U$ contains a ball around $(0,y)$. What does that tell you about $f(U)$?