Suppose $f:A\rightarrow \Bbb R $ is Lipschitz continuous and its derivative $Df$ is continuous on a closed subset $E$, $E\subset A$. Define $$\eta_\delta(a)=\text{sup}_{0<|x-a|<\delta \text{ and }x\in E}\frac{|f(x)-f(a)-Df(a)(x-a)|}{|x-a|}$$. I would like to proof the pointwise convergence for $\delta\rightarrow0$. I started as follows：
By applying Mean Value Theorem, for some $y\in[a,x]\cap E$: $$\frac{|f(x)-f(a)-Df(a)(x-a)|}{|x-a|}=\frac{|(Df(y)-Df(a))(x-a)|}{|x-a|}\\ \leq \frac{|Df(y)-Df(a)||x-a|}{|x-a|}=|Df(y)-Df(a)|$$. Since $Df(x)$ is continuous on $E$, the last term goes to zero when $\delta$ goes to zero.
Question:
The first argument when I used the MVT, is this justified? Since MVT is only formulated on set like $[a,b]$. In my case, this need not to be true since $E$ can be arbitrary closed set, even with empty interior.
How can I proof the claim if I cannot use MVT.