Let $f: \mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}$ be projection onto the first coordinate and let $p$ be the restriction of $f$ to the space $X$ consisting of all points $(x,y)$ with $x\geq 0$ or $y=0$ (or both). Prove that $p$ is a quotient map that neither an open map nor a closed map.
It's obvious that $f$ is continuous and a surjective. In order to prove that $f$ is quotient map, it's enough for us to prove that if $f^{-1}(U)$ is open, then $U$ is open in $R$. I can't find a proper way to explain this. Additionally, I'm also stuck on finding the counterexamples of last question. Could anyone give a hint to move on?
HINT: For the counterexamples consider $[0,1)\times(0,1)$ and the intersection with $X$ of the graph of $y=\frac1x$.
To show that $p$ is a quotient map, let $L=\Bbb R\times\{0\}$, and use the fact that if $p^{-1}[A]=(A\times\Bbb R)\cap X$ is open in $X$ for some $A\subseteq\Bbb R$, then $A\times\{0\}$ is open in $L$.