A compact convex subset $D$ in $\mathbb{R}^n$ with nonempty interior is homeomorphic to $D^n$.
Without loss of generality we may suppose the origin is an interior point of $D$. We claim that the map $\phi: v \mapsto v/\|v\|$ maps the boundary $\partial D$ homeomorphically onto $S^{n-1}$. By convexity, $D$ is homeomorphic to the cone on $\partial D$, and therefore to the cone on $S^{n-1}$ which is $D^n$.
Shouldn’t the claim be obvious?
The restricted map $\phi: \partial D \to S^{n-1}$ is continuous.
It’s surjective: $D$ contains a ball $B = B_{\varepsilon}(0)$ in its interior, and for each nonzero $x \in B$, the positive ray through $x$ intersects $D$ in a bounded half-open line segment. For the extreme point $v$ on this line segment, $\phi(v) = \phi(x)$. Thus every unit vector $u \in S^{n-1}$ is of the form $\phi(v)$ for some extreme point $v \in D$, and such extreme points lie in $\partial D$.
It’s injective: for this we need to show that if $v, w \in \partial D$ are distinct points, then neither is a positive multiple of the other. Supposing otherwise, we have $w = t v$ for $t \gt 1$, say. Let $B$ be a ball inside $D$ containing $0$; then the convex hull of $\{w\} \cup B$ is contained in $D$ and contains $v$ as an interior point, contradiction.
So the unit vector map, being a continuous bijection $\partial D \to S^{n-1}$ between compact Hausdorff spaces, is a homeomorphism.
Any compact convex set $D$ of $\mathbb{R}^n$ is homeomorphic to a disk.
$D$ has nonempty interior relative to its affine span which is some $k$-plane, and then $D$ is homeomorphic to $D^k$ by the theorem.
I guess I’d now like to add a little more to this little article, since it started out being about compact convex sets.
Let $D$ be a compact convex set of $\mathbb{R}^n$. The essential interior of $D$ is the set of points that are interior points of $D$ relative to its affine span. The essential boundary of $D$ is the set of points that do not belong to the essential interior of $D$.
If $x$ belongs to the essential boundary of $D$, then there is some functional $\alpha: \mathbb{R}^n$ such that $\alpha(x) = 1$ and $\alpha(y) \lt 1$ for every essential interior point $y$.
In this case, $\alpha^{-1}(1) \cap D$ is compact, convex, and has dimension strictly less than the dimension of $D$.
(Minkowski-Carathéodory) A compact convex $D$ subset of $\mathbb{R}^n$ is the convex hull of its extreme points. Indeed, every point $x \in D$ is a convex combination of $k+1$ of its extreme points if the affine span of $D$ is $k$-dimensional.
This of course is a baby version of the Krein-Milman theorem.
By induction on $k$. The case for $k=0$ is trivial.
(to be continued)