I suppose $n < m$($m > n$ is same), and the map $f:U \to U'$ is the homeomorphism. We have $\mathbb{R}^n$ is closed in the $\mathbb{R}^m$ then $U$ is closed in $\mathbb{R}^m$. Can we come to the contradiction that if $U'$ is open then $U$ is open too.
But the fact:if $U'$ is open then $U$ is open too only makes sense on the sphere?How can I improve my proof?
As you noted, there's a natural embedding $\mathbb{R}^n \subset \mathbb{R}^m$. Consider the inverse map $f^{-1}: V \to U$. It is also a homeomorphism, and so it's continuous and injective. Since $U \subset \mathbb{R}^n$. we can compose $f^{-1}$ with inclusion map $i: \mathbb{R}^n \hookrightarrow \mathbb{R}^m$ to get a map $i \circ f^{-1}: V \to \mathbb{R}^m$ that is continuous and injective. Since $V \subset \mathbb{R}^m$, invariance of domain tells us that its image is open in $\mathbb{R}^m$. But its image is contained in $\mathbb{R}^n$, which has empty interior in $\mathbb{R}^m$, a contradiction.