How to prove that whether it is a Banach space or not? We consider the Banach space of all continuous functions on $X$ such that for each $f$ in the space, 
\begin{equation*}
||f||=\sup_{x\neq y}\frac{\left\vert
f(x)-f(y)\right\vert }{\left\vert x-y\right\vert }.
\end{equation*}
How I can prove that it is a Banach space? 

Suppose $\{f_n\}$ is a Cauchy sequence in $(CL^0,\|\cdot\|_{CL^0})$. Then


*

*$\{f_n\}$ is bounded (in the $\|\cdot\|_\infty$ norm), and

*$\{f_n\}$ is equicontinuous.


To see the first point, let $M:=\sup_n\|f_n\|_n$, which is finite. For any $n$, one has
$$|f_n(x)|=|f_n(x)-f_n(0)| \le m(f_n)|x-0| \le M|x|\le M,$$
so $\|f_n\|_\infty \le M$. To see the second point, fix $\epsilon>0$ and let $\delta:=\epsilon/M$. Then for all $n$ and all $x,y\in[0,1]$, if $|x-y|<\delta$ then
$$|f_n(x)-f_n(y)| \le m(f_n)|x-y| \le M\delta = \epsilon.$$
It follows from the Arzela-Ascoli theorem that $\{f_n\}$ has a uniformly convergent subsequence, i.e. there exists $\{n_k\}$ and a continuous function $f$ such that $f_{n_k}\to f$ uniformly. Of course, we immediately have $f(0)=0$ and $m(f) \le M$, so $f\in CL^0$. It remains to show $f_{n_k} \to f$ in $\|\cdot\|_{CL^0}$.
Fix $\epsilon>0$. There exists $K$ such that $\|f_{n_j}-f_{n_k}\|_{CL^0} < \epsilon/3$ for all $j,k\ge K$. Now let $x,y\in[0,1]$ with $x\neq y$. Since $f_{n_k}\to f$ uniformly, there exists $K^{x,y} \ge K$ such that $\|f_{n_{K^{x,y}}}-f\|_\infty < \epsilon|x-y|/3$. It follows that, if $k\ge K$,
\begin{align*}
&\frac{|(f_{n_k}-f)(x) - (f_{n_k}-f)(y)|}{|x-y|} \\
&\qquad\qquad\qquad\le \frac{|(f_{n_k}-f_{n_{K^{x,y}}})(x) - (f_{n_k}-f{n_{K^{x,y}}})(y)|}{|x-y|} + \frac{|(f_{n_k}-f)(x) - (f_{n_k}-f)(y)|}{|x-y|} \\
&\qquad\qquad\qquad\le \|f_{n_k}-f_{n_{K^{x,y}}}\|_{CL^0} + \frac{2\|f_{n_k}-f\|_\infty}{|x-y|} \\
&\qquad\qquad\qquad\le \frac{\epsilon}3 + \frac{2\epsilon}3 = \epsilon.
\end{align*}
Crucially, since $K$ does not depend on $x$ or $y$, we can take supremum over $x,y\in[0,1]$ with $x\neq y$ and conclude that, if $k\ge K$,
$$\|f_{n_k}-f\|_{CL^0} \le \epsilon.$$
