Show that the following space is not Hausdorff

Question

Let $S= \{ (x,y) \in \mathbb{R}^2:x=0 \ \text{or} \ y=0 \}$. Let $\mathcal{T}$ be the topology on $ \mathbb{R}^2$ and define $f: \mathbb{R}^2 \to S$ by $f(0,y)=(0,y)$ and $f(x,y)=(x,0)$ if $x \neq 0$. Let $\mathcal{U}$ the quotient topology on $S$ induced by $f$. Show that $(S, \mathcal{U})$ is not Hausdorff. I supose that I have to find two points $(a,b), (x,y) \in S$ with $(a,b) \neq (x,y)$ such that for every open set $U,V$ in $S$ with $(a,b) \in U$ and $(x,y) \in V$ and $U \cap V = \emptyset$, but I tried a lot of times and I didn't find it.

Answer 1

HINT: Show that if $y_0\ne y_1$, then $f(0,y_0)$ and $f(0,y_1)$ do not have disjoint open nbhds in $S$.