Set $X:=\mathbb R^2$. It is to be proven directly that $g:\mathbb R^2\to \mathbb R, g(x,y)=x+y^2$ is quotient map.
$g$ is continuous. $g$ is surjective because for any $r\in \mathbb R, g(r,0)=r$.
It only remains to show the following: let $g^{-1}(U)$ be open in $X$. It's to be shown now that $U$ is open.
Suppose on the contrary that $U$ is not open. So there exists $t\in U$ , which is not an interior point of $U$. It follows that for every $n\in \mathbb N$, there exists an $x_n\in (t-\frac 1n, t+\frac 1n), x_n\notin U$.
By continuity of $g$, $g^{-1}(t-\frac 1n, t+\frac 1n)$ is open. I'm not sure how to get contradiction from here.
Background: I'm trying to do exercise $4 (a),$ section $(22)$ of Munkres' without using previous exercises of the same section as done here.
If $r \colon X \to Y$ is a continuous map, and there is a continuous map $s \colon Y \to X$ such that $r \circ s = {\rm id}_Y$, then $r$ is a quotient map.
Proof: The equality $r \circ s = {\rm id}_Y$ clearly implies that $r$ is surjective. Now, if $U$ is a subset of $Y$ such that $r^{-1}(U)$ is open in $X$, then $U$ is open in $Y$: By the continuity of $s$, the set $$s^{-1}(r^{-1}(U)) = (r \circ s)^{-1}(U) = {\rm id}_Y^{-1}(U) = U$$ is open in $Y$.
In your case, $g \circ i = {\rm id}_{\Bbb R}$ where $i \colon \Bbb R \to \Bbb R^2$ is defined by $i(x) = (x,0)$.