 Solution

Let $E=\mathrm{aff}K$ be the affine hull and put $k=\dim E\le n$. If $k=0$, $K$ is a singleton and is homeomorphic to ${\mathbb{B}}^0$. Suppose $k>0$. Choose $k+1$ affinely independent points of $K$ spanning $E$. Their simplex is contained in $K$ because $K$ is a convex set, and its barycentre $a$ has a neighbourhood in $E$ contained in that simplex. Identifying $E-a$ with ${\mathbb{R}}^k$, put $D=K-a$. For some $0<r\le R<\infty$,Here the inner inclusion follows from that neighbourhood, and the outer one follows from the compactness of $K$.

Use the Minkowski functionalThe inclusions imply $|z|/R\le p(z)\le |z|/r$. The fact that $D$ is a convex set implies $p(z+w)\le p(z)+p(w)$: if $z\in sD$ and $w\in tD$, then $(z+w)/(s+t)$ is a convex combination of points of $D$. Also $p(tz)=tp(z)$ for $t\ge 0$. ConsequentlyThus $p$ is a continuous function. Since $D$ is closed and contains zero, $D=\{p\le 1\}$: the forward inclusion is immediate, and for the reverse inclusion one takes a limit of $z/t_j\in D$ with $t_j\to p(z)\le 1$, using positive homogeneity and the fact that $D$ is a convex set. For $z\ne 0$, $p(z)>0$, so this limiting argument has no division by zero.

Define a radial continuous map byEach ray meets $D$ in exactly $\{s\theta :0\le s\le 1/p(\theta )\}$. Hence $h$ is a bijection from ${\mathbb{B}}^k$ to $K$. Away from zero its continuity follows from that of $p$, and at zero $|h(t\theta )-a|\le Rt$. A continuous map that is a bijection from a compact space to a Hausdorff space is a homeomorphism. ThereforeUsing the affine hull is essential: $K$ can have empty interior in the ambient ${\mathbb{R}}^n$.