Theorem:Suppose $U \subset \mathbb{R}^n, V \subset \mathbb{R}^m$ are open and $f: U \to V$ is a diffeomprhism with inverse $g : V \to U$. Then
(i) $n=m$, and
(ii) at each $x \in U$ the differentials are invertible linear maps with
$dg_y=(df_x)^{-1}$ or $J_g(y)=(J_f(x))^{-1}$ for $y:=f(x)$.
Proof: We differentiate $id_U=g \circ f$:
$id_{\mathbb{R}^n}=d(id_U)_x=d(g \circ f)_x=dg_{f(x)} \circ df_x$.
Similarly, $id_{\mathbb{R}^m}=df_x \circ dg_{f(x)}$. Thus $dg_{f(x)}$ is invertible with (left and right) inverse $df_x$. This proves (ii) and since $dg_{f(x)}$ is invertible we have $n=m$ which proves (i).
Now my question is why we have assumed that $f$ is a diffeomorphism since we did not use the continuity of the derivative anywhere. I think the same proof should work for a differentiable homeomorphism with a differentiable inverse. The lecture notes also mention that homeomorphism preserve dimension in general, but since the notes lead up to the inverse mapping theorem I am not interested in a general proof here.
I know this might be a simple question, but I am pretty new to homemorphisms and diffeomorphism etc., so I want to make sure I don't misunderstand anything.
Thanks for any help!