Entire Question: Let $S\subseteq \mathbb{R}^n$ be an open set and $f:S\to \mathbb{R}^m$ be a $C^1$ function. We say that $f$ is $C^1$ on $\overline{S}$ if for every $p\in \overline{S}$, there is an open neighborhood $U_p\subseteq \mathbb{R}^n$ of p and a $C^1$ function $\tilde{f}:U_p\to \mathbb{R}^m$ s.t $f\vert_{S \cap U_p}=\tilde{f}\vert_{S \cap U_p}$. Show that $D\tilde{f}(p)$ is independent of the choice $\tilde{f}$ and $U_p$, and therefore there is no ambiguity in saying $Df$ is the derivative of $f$ on $\overline{S}$.
I don't believe I've proven this correctly. It seems to be a question aimed at defining differentiability on a closed set. And what I got was showing differentiability on a larger open set.
Fix some arbitrary $p\in \overline{S}$. Let $U_p\subseteq \mathbb{R}^n$ be some open neighborhood of $p$. Then $S \cap U_p\subseteq S$ and $S \cap U_p\subseteq U_p$. Since $f$ is $C^1$ there is a $C^1$ function $\tilde{f}:U_p\to\mathbb{R}^m$ such that $\tilde{f}\vert_{S\cap U_p}=f\vert_{S\cap U_p}$. Let $g=f\vert_{S\cap U_p}$ and $\tilde{g}=\tilde{f}\vert_{S\cap U_p}$.
Since $f,\tilde{f}$ are $C^1$, $Dg(x)=Df(x)$ and $D\tilde{g}(y)=D\tilde{f}(y)$ for all $x,y\in S \cap U_p$ and these derivatives exist for each $x,y$. Then for some $a\in S\cap U_p$ $$\lim\limits_{\vert h\vert\to 0}\frac{\vert \tilde{g}(a+h)-\tilde{g}(a)-D\tilde{g}(a)(h)\vert}{\vert h\vert}=0.$$ But since $g=\tilde{g}$, $$\lim\limits_{\vert h\vert\to 0}\frac{\vert \tilde{g}(a+h)-\tilde{g}(a)-D\tilde{g}(a)(h)\vert}{\vert h\vert}=\lim\limits_{\vert h\vert\to 0}\frac{\vert g(a+h)-g(a)-D\tilde{g}(a)(h)\vert}{\vert h\vert}=0.$$ But derivatives are unique, $Dg(a)$ is the unique linear approximation satisfying this limit. So $Dg(a)=D\tilde{g}(a)$ for all $a\in S \cap U_p$ and since $Df(x)=Dg(x)$ and $D\tilde{f}(y)=D\tilde{g}(y)$ for all $x,y \in S \cap U_p$ and then $Df(x)=Dg(x)=D\tilde{g}(x)=D\tilde{f}(x)$ for all $x\in S \cup U_p$. Where $S \cup U_p$ is an open set and $\overline{S}\subseteq S \cup U_p$
Thus $f:S \cup U_p \to \mathbb{R}^m$ is $C^1$ and so $Df$ is the derivative of f on $\overline{S}$.
Looking at it now I can see I've definitely gone wrong here at least.
$Df(x)=Dg(x)=D\tilde{g}(x)=D\tilde{f}(x)$ for all $x\in S \cup U_p$. Where $S \cup U_p$ is an open set and $\overline{S}\subseteq S \cup U_p$