It is much easier to think about this in polar coordinates since a point on the unit circle can be uniquely identified by an angle $\theta\in (0,2\pi]$. So we identify $S^1$ with the space $(0,2\pi]$ under the topology given by "circular intervals" that possibly wrap around $2\pi$. (There is an explicit homeomorphism $\theta\mapsto (\cos \theta,\sin\theta)$ that you can use to translate back and forth.) It's more convenient to use $2\pi$ for "angle $0$" in this case.
Then you can view $S^1/\mathscr{R}$ as $(0,\pi)\cup\{\ast\}$ where $\ast$ is the single point corresponding to the class of angles in $[\pi,2\pi]$.
The map your partner suggests (modulo an issue discussed below) is $f:S^1/\mathscr{R}\to S^1$ where $f(t)=2t$ for $0<t<\pi$ and $f(\ast)=2\pi$. Described out loud, this map doubles all angles in the upper half circle $(0,\pi)$ so that they cover everything in $(0,2\pi)$, and then sends $\ast$ to $2\pi$ to fill the last point.
Written this way, it is quite clear that the map is a bijection. Continuity is not hard to prove either. A basic open set is a set of the form $(r,s)$ or $(0,r)\cup (s,2\pi]$ for $0<r<s\leq 2\pi$. In the first case the preimage is $(r/2,s/2)$. In the second case the preimage is $(0,r/2)\cup (s/2,\pi)\cup\{\ast\}$ (which is open in the quotient topology since its preimage under the quotient map back to $S^1$ is $(0,r/2)\cup (s/2,2\pi]$.)
I will add some further details: I have identified $(0,2\pi]$ and $S^1$ to make the problem easier. Formally, the underlying map in this identification is $\theta\mapsto (\cos\theta,\sin\theta)$. I am not saying that this is the homeomorphism you ask for in the main question. The homeomorphism is the map $f$ which is the same as your partners (modulo issue) up to this identification between $S^1$ and $(0,2\pi]$.
If you want to unwrap the identification to get a function in Cartesian coordinates then you just have to go through the trig. Start with $(x,y)$ in the upper half circle. Go polar by taking $\theta=\arccos(x)$ (which is in $(0,\pi)$). Apply $f$ to get $2\arccos(x)$. Now go back to Cartesian coordinates to get $(\cos(2\arccos(x)),\sin(2\arccos(x))$. We have our function.
Actually, going through this I can see either there is a typo or your partner didn't quite get it right: $\arccos(y)$ should be $\arccos(x)$. If it looks strange to have the map depend only on $x$ then recall that a point in the upper half of the unit circle is completely determined by its $x$ coordinate. So really this is a "one parameter" situation, which is all the more reason to go polar.
Yes 2 is equivalent to $f^{-1}[\{y\}], y \in Y$ being equal to the partition of $X$ into classes. This both shows 1-1-ness of $\tilde{f}$ (if two classes map to the same $y$ under $\tilde{f}$, they were already in the same class) as well as well-definedness (if two points are from the same class, they have the same $f$ image).
In your $\Bbb S^2, \mathbb{ RP}^2$ example
the universal property of the topology implies that if $f$ is continuous, so is $g$ (as $g = f \circ q$), so the work is in showing $f$ continuous. $2$ is easy as two distinct points on the sphere define the same line iff they are antipodal. That fact, plus that every class is obtained, makes $g$ a bijection. You still have the show continuity of $f$ though.
Note that in this case the last two steps are unnecessary as a continuous bijection from a compact space to a Hausdorff space is already a homeomorphism.
Best Answer
The map $f$ should be from $\Bbb S^1/(x \sim -x)$ to $\Bbb S^1$, given by $f([z]) = z^2$. Note that $[z] = \{z,-z\}$ for every $z\in \Bbb S^1$. $f$ is well-defined, for if $[z_1] = [z_2]$, then either $z_1 = z_2$ or $z_1 = -z_2$. In any case, $z_1^2 = z_2^2$, so $f([z_1]) = f([z_2])$.
We show that $f$ is a continuous bijection. Since $\Bbb S^1/(x\sim -x)$ is compact, and $\Bbb S^1$ is Hausdorff, $f$ is a homeomorphism.
I hope you can show the continuity of $f$ to conclude.