A nonempty open subset of $\mathbb{R}^n$ is not diffeomorphic to a nonempty open subset of $\mathbb{R}^m$ , for $m\neq n$. To solve this problem, let $U$ be an open subset of $\mathbb{R}^n$, $V$ an open subset of $\mathbb{R}^m$. Suppose there exists $f:U\to V$ that is diffeomorphism.
I think I need to show that the map $f'(p):\mathbb{R}^n\to \mathbb{R}^m$ is linear isomorphism. But I have no idea.
The comment Eric Towers posted is the gist of it, but I'm providing the details. I suppose that you're not seeing this in a manifold-orientated course/ book since in that case you'd already know the answer.
So Let's suppose that $f$ is a diffeomorphism and $g$ is its inverse, namely $g=f^{-1}$. Then define the differential maps: $f'_p:\mathbb{R}^n\to\mathbb{R}^m$ $g'_{f(p)}:\mathbb{R}^m\to\mathbb{R}^n$. These are linear maps and by the chain rule it is $(g\circ f)'_p=g'_{f(p)}\circ f'_p$. But it is $g\circ f=Id_{U}$ since $g=f^{-1}$. Hence $(g\circ f)'_p=Id_{\mathbb{R}^n}$. Therefore $g'_{f(p)}, f'_p$ are linear isomorphisms between $\mathbb{R}^n$ and $\mathbb{R}^m$. Is this possible for $n\neq m$?