Open sets in Quotient topology 
Two points are equivalent $(x_1,y_1)\sim(x_2,y_2)\iff (x_1,y_1)=t(x_2,y_2)$ for $t>0$. Let $Y:=\frac{\mathbb{R}^2}{\sim}:=\{[(x,y)]:(x,y)\in\mathbb{R}^2\}$ denote the quotient space under the quotient topology induced by $\pi:\mathbb{R}^2\to Y$ by $\pi(x,y)=[(x,y)]$. Which of the following statements are true regarding the space $Y$?
$a)$ $Y$ is $T_1$ but not $T_2$.
$b)$ $Y$ is neither $T_1$ nor $T_2$.
$c)$ $Y$ is compact.
$d)$ $Y$ is not compact.

As per the definition of quotient topology, $U\subset Y$ is open in $Y$ if $\bigcup_{[x]\in U} [x]$ is open in $X$. I can see that $\varnothing,Y$ are open in $Y$. But I am unable to figure out if there are any other open sets.
Also graphically, we can see that the equivalence classes are lines passing through the origin and the point $(x,y)$. Lines are closed in $\mathbb{R}^2$ so any finite collection of them are closed too. If we take any countable collection of these lines then its complement will never be closed since the origin being a limit point of the complement is never in it but not open either. 
While the countable collection of these equivalence classes will be closed since "$S\subset \mathbb{R}^n$ is closed iff for every $x\in S$ every $n$-ball $B(x)$ intersects $S$ in atleast one point different from $x$ in $S$."
My question here is what are the open sets in this quotient topology?
 A: The equivalence classes of $Y$ are the origin and the open half lines with the origin for extremity. Denote by $\pi$ the projection from $\mathbb R^2$ onto $Y$.
$Y$ is not $T_1$ as an open half line is not closed in $\mathbb R^2$ (see Quotient space (topology)).
$Y$ is not $T_2$ as the origin and a half line can't be separated in $\mathbb R^2$.
$Y$ is compact. Consider an open cover $\mathcal U=\{U_i \mid i \in I\}$ of $Y$. By definition $V_i=\pi^{-1}(U_i)$ is an open subset of $\mathbb R^2$ for all $i \in I$. Denote $\mathcal V=\{V_i \mid i \in I\}$.
The circle $C \subset \mathbb R^2$ centered on the origin with radius equal to $1$ is compact in $\mathbb R^2$. Hence it can be covered with a finite subset $\mathcal V_1=\{V_i \mid i \in I_1\}$. As $\pi$ is onto, the finite set $\mathcal U_1=\{U_i \mid i \in I_1\}$ covers $Y$ with potentially $U_{i_0}$ such $\{(0,0)\} \in U_{i_0}$.
A: Write points of $\mathbb{R}^2$ in polar coordinates $(\rho,\theta)$. For $\alpha, \beta$, consider the open arc $U_{\alpha,\beta} := \{[(1,\theta)] \big| \ \alpha < \theta < \beta \} $. 
The topology on $Y$ has the following basis of open sets : (edit)
 $\{ U_{\alpha,\beta} \big| \ \alpha,\beta \in \mathbb{R}\} \bigcup \{\{Y\}\}$

Edit Since the image of every neighborhood of $(0,0)$ under the quotient map is $Y$, $Y$ is the only neighborhood of $[(0,0)]$.

