Let $X$ be a Banach space. Show that $L=\{f:X\to\mathbb{R}: f \mbox{ is Lipschitz}, f(0) = 0\}$ with the norm

$$||f||\_{Lip\_0} = \sup\left\{\frac{|f(x)-f(y)|}{||x-y||}, x\neq y\in X\right\}$$

is a Banach space.

I've found [Banach space of p-Lipschitz functions](https://math.stackexchange.com/questions/1213529/banach-space-of-p-lipschitz-functions) but I did not understand the proof given. 

I have a few questions first. Which norm is $||x-y||$? 

So I need to prove that every Cauchy sequence in $L$ converges to an element of $L$, right?

In other words, $\forall \epsilon>0$ there exists $n\_0$ such that $m,n>n\_0\implies ||f\_m-f\_n||\_{Lip\_0}<\epsilon$

$$ ||f\_m-f\_n||\_{Lip\_0} = \sup\left\{\frac{|(f\_m-f\_n)(x)-(f\_m-f\_n)(y)|}{||x-y||}, x\neq y\in X\right\} = \sup\left\{\frac{|f\_m(x)-f\_m(y)|}{||x-y||}+\frac{f\_n(y)-f\_n(x)}{||x-y||}, x\neq y\in X\right\}$$

both $f\_m$ and $f\_n$ are Lipschitz so they're continous, which means something I don't know what.