Loring W. Tu defines a smooth function from an arbitrary subset as:
Let $\ S \subset \mathbb{R}^n$ be an arbitrary subset. A function $\ f: S \to \mathbb{R}^m $ is smooth at a point p in S if there exist a neighborhood U of p in $\ \mathbb{R}^n $ and a smooth function $\ g: U \to \mathbb{R}^m $ such that $\ f = g $ on $\ U \cap S$ The function is smooth on S if it is smooth at each point in S.
Theorem: 22.3 Smooth invariance of domain
let $\ U \subset \mathbb{R}^n$ be an open subset, $\ S \subset \mathbb{R}^n$ an arbitrary subset, and $\ f: U \to S $ diffeomorphism. Then S is open in $\ \mathbb{R}^n$
Is this proof valid?
For every point $\ p \in S$,$\ g_p$ is a smooth function on some neighborhood $\ V_p$ of p, such that $\ g_p=f^{-1} $ on $\ V_p \cap S $, by definition given above.
Set $\ g(x)= \sum \rho (x)g_p(x) $, where $\ \rho$ is the partition of unity subordinate to $\ V_p$. $\ g$ is smooth and equals $\ f^{-1} $ at every point of S. (I think this part is pretty obvious, but tedious, so I omitted the proof of it)
$\ g$ maps an open subset $\ \bigcup V_p$ to a subset containing $\ U$. Since it is smooth, its inverse sends open subsets to open subsets. $\ g^{-1}(U) = S$. $\ S$ is open.