Here's the example in question for reference:
We have to define a map $X^* \to S^2$ that's a homeomorphism. We think of the points in $X$ in their polar form $(r, \theta)$ and those in $S^2$ in cylindrical form $(\rho, \theta, z)$. Then we define $g: X \to S^2$ by $g(r,\theta)=(\sqrt{1-(2r-1)^2}, \theta, 2r-1)$. Then it's easy to see that $g$ is bijective and continuous.
I'm stuck now with how to proceed after this. I clearly need another bijective continuous map $h:X^* \to X$ defined in a way that gives $g\circ h$ the required homeomorphism. I tried defining $h$ as $h(S^1)=(1,0)$ and $h(\{x\times y\})=x\times y$ for $x\times y \in X\setminus S^1$.This is clearly not bijective and I'm not even sure what to say about it's continuity. I don't think a map like this will work because of the problematic choice of choosing an image of $S^1$.
Please note that I've consulted a bunch of questions on this site asking about this same example and my approach borrows heavily from the answers to those posts.
Edit: Or, if I define a map $f:X^* \to S^2$ directly by $f(\{0\times 0\})=(0,0,-1), f(S^1)=(0,0,1)$ and for all other $\{x\times y\} \in X^*$ we define $f(\{x\times y\})=f((r, \theta))= (\sqrt{1-(2r-1)^2}, \theta, 2r-1)$, then this looks like a bijective function. I'm not sure how should I go about proving this gives me a homeomorphism. In another post it was mentioned that $X^*$ is compact and if that's so then just proving continuity will suffice to prove that $f$ is a homeomorphism (because $S^2$ is Hausdorff). I'm not sure how to prove that $X^*$ is compact, or that $f$ is continuous.
I've found a result which proves that $X^*=X/S^1$ is compact. All we need is that $S^1$ be identified with a point (which it is) and we need $S^1$ to be a closed subspace of $X$ (which it is being a compact subspace of a Hausdorff space). All that remains is to prove that $f$ is continuous.
A couple of useful tricks of the trade. As you mention, a continuous bijection from from a compact space to a Hausdorff space is a homeomorphism. Why is $X^*$ compact? Because it's the continuous image of the compact space $X$. Why is $f$ continuous? Because (writing $q$ for the quotient map $q:X\to X^*$) a map $f$ out of $X^*$ is continuous if and only if the composition $f\circ q$ is continuous.
In a similar spirit, $f\circ q : X \to S^2 \subseteq {\Bbb R}^3$ is continuous (by the characterizations of maps into subspaces and products) if and only if each component of $q\circ f:X\to {\Bbb R}^3$ is continuous.