I've got this question on my mind for a while and I can't find an argument for it. Here goes my question.
Let $U\subset\Bbb R^n$ an open set and $f:U\to\Bbb R$ differentiable at $a\in U$. Exist $r>0$ such that $B(a,r)\subset U$, then for every vector $h\in\Bbb R^n$ with $|h|<r$ follows $a+h\in B(a,r)$. By the differentiability at $a$ we have
$$f(a+h)-f(a)=\langle\nabla f(a),h\rangle+r(h),\quad\lim_{h\to0}\frac{r(h)}{h}=0.$$
Since $\displaystyle\lim_{h\to0}\frac{r(h)}{h}=0,$ exist $\varepsilon>0$ such $|h|<\varepsilon\Rightarrow|r(h)|<|h|$. Then, for every $|h|<\varepsilon$ we have $$|f(a+h)-f(a)|=\left|\langle\nabla f(a),h\rangle+r(h)\right|\leq\left|\langle\nabla f(a),h\rangle\right|+|r(h)|\leq|\nabla f(a)||h|+|h|.$$ So we get $M>0$ such that $|f(a+h)-f(a)|\le M|h|$.
But why is it not possible tweak these steps to find $\varepsilon>0$ and $M>0$ such that $|f(x)-f(y)|\leq M|x-y|$ for every $x,y\in B(a,\varepsilon)$ ?
The easy answer: There are counterexamples. For example, suppose we take $n=1$ and define $f(x)=x^2$ if $x$ is rational, $f(x)=0$ otherwise. Then $f'(0)=0.$
But for any $r>0$ you can find a non-zero rational $x\in (-r,r).$ For this $x,$ there is a sequence $y_n$ of irrationals converging to $x.$ For this sequence we have
$$\tag 1 \frac{f(y_n)-f(x)}{y_n-x}=\frac{-x^2}{y_n-x}.$$
This is unbounded as $n\to\infty$ because the numerator just sits there at a nonzero value, while the denominator $\to 0.$ And because $(1)$ is unbounded, we have a counterexample.