Is this right? Showing that if $S\subset \mathbb{R}^n$ arbitrary and $f$ a $C^r$ function then there is $A$ open such that $f$ is $C^r(A)$.

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Let $f : S \to \mathbb{R}$ a $C^r$ function, where $S$ is any subset of $\mathbb{R}^n$. We say that $f$ is differentiable on $x_o \in S$ if there is $U_{x_0}$ and $g : U_{x_0} \to \mathbb{R}$ such that $g$ is differentiable in $x_0$ and $g = f$ on $U_{x_0} \cap S$.

The result I am trying to prove is the following:

Let $f$ as stated. Then there is $A$ open set containing $S$ such that $f$ can be extended to a $C^r(A)$ function.

My attempt:

Take $x \in S$. Then there is $g_x$ such that $f = g_x$ on $U_x\cap S$ and $g_x$ is differentiable on $U_x.$ Take this for each $x \in S$. Then $\{U_x\}$ is an open cover for $S$. Now take an unity partition associated to this partition, $\{\rho_x\}$. Make $h(y) = \sum_{x\in S} \rho_xg_x(y)$ and $A = \cup_{x\in S}U_x.$

The I claim that this is the function searched.

Indeed, if $y \in S$ then $h(y) = \sum_{x\in S} \rho_xf(y) = f(y).$