Definition: Let $S$ be an arbitrary subset of a smooth manifold $M$. Let $f:S \to \mathbb{R}$ be a map. The map $f$ is said to be $C^r$ if for every $s \in S$ there is an open set $O \subset M$ containing $s$ and a map $\tilde{f}$ that is $C^r$ on $O$ and such that $\tilde{f}_{|S \cap O} = f$.
I am trying to prove that if a function is smooth (with the definition above) on a set $S \subset M$, then it has a smooth extension to an open set that contains $S$.
I am aware that the OP of this post Extension of a smooth function on a set of a manifold to an open nbd of the set. gives the following justification:
Assume $f:S→\mathbb{R}$ is smooth and for each $s∈S$ let $U_s$ be a nbd of $s$ and $f_{s}:U_s→\mathbb{R}$ be a smooth map such that $f_s{_{|U_s \cap S}}=f$. Let $\{ϕ_s\}_{s∈S}$ be a partition of unity dominated by $\{U_s\}_{s∈S}$. For each $x∈X$ we define $A_x=\{s∈S:x ∈ supp(ϕ_s)\}$. We define $g:⋃_{s∈S}U_s→\mathbb{R}$ as $g(x)=∑_{s∈A_x}ϕ_s(x)f_s(x)$. Then g is the desired smooth extension of f.
It is unclear to me how to obtain this partition of unity since the family $\{U_S\}_{s \in S}$ is not a cover for $M$.
The family $\{U_s\}_{s \in S}$ is an open cover of $\bigcup_{s\in S} U_s,$ which is an open subset of a manifold and thus a manifold.