Here is an example: $N=1$, $E=[0,e^{-3})$ and
$$f(x)=\left\{\begin{array}{ccc}(\log x)^{-2}&,& x\in(0,e^{-3})\\ 0&,&x=0\end{array}\right..$$
For every $\alpha>0$,
$$\lim_{x\to 0^+}\frac{|f(x)-f(0)|}{x^\alpha}=\infty,$$
so $f$ is not $\alpha$-Hölder continuous.
Now let us show that $f$ is Dini continuous. To begin with, note that
$$f'(x)=-\frac{2}{x(\log x)^3}>0,\quad x\in(0,e^{-3}),$$
and
$$f''(x)=\frac{2}{x^2(\log x)^3}+\frac{6}{x^2(\log x)^4}<0,\quad x\in (0,e^{-3}).$$
Therefore, $f$ is increasing on $[0,e^{-3})$ and $f'$ is decreasing on $(0,e^{-3})$. As a result, for every $t\in (0,e^{-3})$, when $0\le x<y\le x+t<e^{-3}$,
$$0\le f(y)-f(x)\le f(x+t)-f(x)=\int_0^tf'(x+s)ds \le \int_0^tf'(s)ds=f(t).$$
It follows that
$$\omega_f(t)\le f(t),\ \forall t\in (0,e^{-3})\Longrightarrow \int_0^{e^{-3}}\frac{\omega_f(t)}{t}dt\le \int_0^{e^{-3}}\frac{f(t)}{t}dt<\infty.$$
It doesn't feel quite right.
And rightly so. Consider for example (identifying $\mathbb{C}$ with $\mathbb{R}^2$) a slit annulus,
$$D = \{ z \in \mathbb{C} : 1 < \lvert z\rvert < 2\} \setminus (1,2),$$
and $u(z) = \arg z$. Then the derivative of $u$ is Hölder-continuous, but $u$ is not Lipschitz continuous.
You can of course construct similar examples in higher dimensions.
What you need for the conclusion is that the distance of two points in $D$ doesn't differ by too much from the length of the shortest paths in $D$ between the points.
Best Answer
Check this paper and the first reference paper. https://arxiv.org/pdf/1407.6871.pdf