Let $U \subset \mathbb{R}^n$ be an open set. Suppose that the map $h:U \to \mathbb{R}^n$ is a homeomorphism from $U$ onto $\mathbb{R}^n$, which is uniformly continuous. Prove $U = \mathbb{R}^n$.
My first attempt guided by intuition was to look at covering $\mathbb{R}^{n}$ by balls of radius $=\frac{1}{n}$ and conclude something forwarding to contradiction by looking at inverse-image of such covering, but I cannot see it for now.
On
Here's a sketch of another argument along the line of Daniel Fischer's.
By the connectedness of $\mathbb{R}^n$, it suffices to show $U$ is closed. So suppose $x_n \in U$ and $x_n \to x$. Then $\{x_n\}$ is Cauchy, and it follows from the uniform continuity that $\{f(x_n)\}$ is Cauchy as well. $\mathbb{R}^n$ is complete, so $f(x_n)$ converges to some $y \in \mathbb{R}^n$. By the continuity of $f^{-1}$, we have $x_n = f^{-1}(f(x_n)) \to f^{-1}(y)$ so $x = f^{-1}(y)$. In particular, $x \in U$.
This would work just as well if we replaced $\mathbb{R}^n$ by another connected complete metric space.
$\mathbb{R}^n$ is a complete metric space. A uniformly continuous map from a metric space $X$ to a complete metric space can be extended to a uniformly continuous map from the completion of $X$ to $Y$.
If $U \neq \mathbb{R}^n$, the extension $\overline{h} \colon \overline{U} \to \mathbb{R}^n$ could not be injective, and that would contradict the assumption that $h\colon U \to \mathbb{R}^n$ is a homeomorphism.