Show that the interior of a non-degenerate convex polygon in $\mathbb{R}^2$ is homeomorphic to the open unit disk in $\mathbb{R}^2$.
My attempt: Let $P$ be the set of points in the polygon, let $c \in \text{int}(P)$. Consider arbitrary $x \in \text{int}(P)$, $x \neq c$. Let $R := \{c+t(x-c): t > 0\} \cap P$. Let $b := \sup_{y \in R} d(y, c)$. Define \begin{eqnarray*} f(x) := \frac{d(x,c)}{b}(x-c). \end{eqnarray*} Finally let $f(c) = (0,0)$.
The problem: while I am pretty sure this is the correct map, I am a hard time proving continuity as I cannot see how to get explicit formulas without knowing which polygon I am mapping to the open unit disk.
P.S. I've seen questions very close to this one, but all answers stopped short of explaining how to show continuity of the map.
First, your quantity $b$ is not a constant, it depends on the ray $R$, and so ostensibly $b$ depends on $x$. But it should be clear that $b=b(\theta)$ really only depends on the angle $\theta$ of the ray $R$. Let me denote $R(\theta)$ as the ray of angle $\theta$ emanating from $c$, and $p(\theta)$ as the point on $R(\theta) \cap P$ furthest from $c$, so $b(\theta) = |p(\theta)-c|$.
The key is to prove that $b(\theta)$ is continuous at each $\theta_0$. Continuity of $f(x)$ away from $c$ then follows, because $\theta = \theta(x)$ has an obviously continuous formula near each $x \ne c$, involving the arctangent function.
By convexity, there exists a closed half-plane $H$ with boundary line $L$ containing $p(\theta_0)$ such that $P \cap H \subset L$. Also, $c \not\in H$ by non degeneracy. Let $\overline H$ be the opposite half-plane to $H$ with the same boundary line $L$. It follows that for angles $\theta'$ near $\theta_0$ we have $p(\theta') \in \overline H$. From this it follows that $b(\theta)$ is upper semicontinuous.
Consider any angle $\theta'$ near $\theta_0$ (any $\theta'$ not pointing opposite $\theta_0$ will do). By convexity, for each $q \in \overline{p(\theta_0) p(\theta')}$ the interior of the segment $\overline{cq}$ must be contained in $P$. From this it follows that $b(\theta)$ is lower semicontinuous.