Let $\Omega \subset \mathbb R^n$ open and $$\varphi:\Omega \to \mathbb R^m$$ continuous and injective. We suppose $m>n$. If $B\subset \bar B\subset \Omega $ is a ball, why $$\varphi|_{\bar B}: \bar B\to \varphi(\bar B)$$ is a homeomorphism ?
It's clear that $\varphi:\bar B\to \varphi(\bar B)$ is continuous and bijective. But why is it open ?
This is because the image of a compact under a continuous map is a compact.
A closed subset of a closed ball is compact (in $\mathbb R^n$). Hence the image of a closed subset is closed. Proving that the restriction of $\varphi$ to $\overline B$ is closed and therefore open as this restriction is a bijection.