We note $E=\Biggl\{f:[0,1]\to\mathbb{R},~\sup\limits_{x\neq x'}\frac{\vert f(x)-f(x')\vert}{\vert x-x'\vert^\frac{1}{2}}<\infty\Biggr\}$
I have showed that $(E,\Vert\cdot\Vert_E)$ is a Banach space with $$ \Vert f\Vert_E = \vert f(0)\vert+\sup\limits_{x\neq x'}\frac{\displaystyle\vert f(x)-f(x')\vert}{\displaystyle\vert x-x'\vert^\frac{1}{2}} $$
Now, I have to show that $$ id\left\{ \begin{array}{lcl} \displaystyle E&\to& \mathcal{C}([0,1],\mathbb{R})\\ \displaystyle f&\mapsto& f \end{array} \right. $$ is compact, where $\mathcal{C}([0,1],\mathbb{R})$ denotes the set of continuous functions from $[0,1]$ to $\mathbb{R}$.
My try: Considering a sequence $(f_n)_{n\in\mathbb{N}}$ that is uniformly $\Vert\cdot\Vert_E$-bounded, I can show that it implies that it is uniformly bounded for $\Vert f\Vert_{\mathcal{C([0,1],\mathbb{R})}}=\sup\limits_{x\in[0,1]}\vert f(x)\vert$ as $\Vert\cdot\Vert_{\mathcal{C([0,1],\mathbb{R})}}\leqslant\Vert\cdot\Vert_E$, but I do not know how to proceed next.
If you could indicate the steps to follow, that would be great, thanks !