The statement in the title of the question is used in the proof of Theorem 2.1 in the page for balls on nLab. I am not sure if we also need the set to be compact.
Theorem 2.1 A compact convex subset $D$ in $\Bbb{R}^n$ with nonempty interior is homeomorphic to $D^n$.
Proof. Without loss of generality we may suppose the origin is an interior point of $D$. We claim that the map $\phi:v\mapsto v/\lVert v \rVert$ 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 $D^n$.
The claim reduces to the following three steps.
- The restricted map $\phi:\partial D\to S^{n-1}$ is continuous.
- It's surjective: $D$ contains a ball $B=B_{\epsilon}(0)$ in its interior, and for each $x\in B$, 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 $w=tv$ for $t>1$, 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. ▮
I know the construction of a cone. If $X$ is a topological space, cone on $X$ is $(X \times I) / (X \times \{0\})$ where I is the unit interval and $/$ is taking quotient, "cylinder $X \times I$ collapsed at one end".
Can someone help me prove the statement with simpler concepts such as compactness, connectedness, closedness, convexity but not contractibility?