HINT: Fix $c\in\Bbb R$, and look at the set of points $\langle x,y\rangle\in\Bbb R^2$ such that $x+y^2=c$: that’s one equivalence class. You can even think of it pictorially: it’s the graph of $x=c-y^2$.
Lemma. Let $X,Y$ be topological spaces and $f:X\to Y$ a continuous surjection. Consider $\sim$ on $X$ given by: $x\sim y$ iff $f(x)=f(y)$. Then $f$ induces the following continuous bijection:
$$F:X/\sim \to Y$$
$$F([x]_\sim)=f(x)$$
which is a homeomorphism if $f$ is a quotient map.
Proof. $F$ is of course well defined. It is easy to see that it is a bijection as well. Furthermore it is continuous because if $\pi:X\to X/\sim$ is the standard projection then $\pi^{-1}(F^{-1}(U))=f^{-1}(U)$ and thus $F^{-1}(U)$ is open.
Finally if $F^{-1}(U)$ is open, then by the above so is $f^{-1}(U)$ and thus if $f$ is a quotient map, then $U$ is open. Meaning $F$ is a quotient map as well. But a bijective quotient map is a homeomorphism. That's because for bijections $V=F^{-1}(F(V))$, and since $F$ is a quotient map, then this implies it is open. Or equivalently its inverse is continuous. $\Box$
With that consider
$$f:\mathbb{R}^n\to [0,\infty)$$
$$f(v)=\lVert v\rVert$$
This function is of course continuous.
Furthermore it is a quotient map. Indeed, assume that $f^{-1}(U)$ is open while $U\subseteq [0,\infty)$ is not. This means that there is $r\in U$ and a sequence $(x_n)\subseteq [0,\infty)\backslash U$ convergent to $r$. Now choose a vector $v$ in $\mathbb{R}^n$ with $\lVert v\rVert=r$. Of course $v\in f^{-1}(U)$. Next take $v_n=\frac{x_n}{r}v$ and note that $\lVert v_n\rVert = x_n$. Finally $v_n\to v$. However $v_n\not\in f^{-1}(U)$ which cannot happen since the last set is open. Note that the special case when $r=0$ has to be treated separately, which I leave as an exercise. Also note that the same works for any dimension (even infinite) and any norm.
Thus, by our lemma, $f$ induces a homeomorphism between $\mathbb{R}^n/\sim$ and $[0,\infty)$. As the final step pick your favourite homeomorphism $[0,\infty)\to [0,1)$, e.g. $\text{arctan}$ or something.
Best Answer
Let $S^1=\{z\in\mathbb{C}\mid|z|=1\}$. This is a common definition for $S^1$ but may not be the one you've been given.
Let $f\colon\mathbb{R}/{\sim}\rightarrow S^1$ be given by $f([t])=e^{2\pi it}$.
Can you show that this is well defined? (That is, show that if $t_1\in[t]_{\sim}$ and $t_2\in[t]_{\sim}$ then $e^{2\pi it_1}=e^{2\pi it_2}$).
Can you show that this is a homeomorphism?