Show that collection of Lipschitz functions with Lipschitz norm is a Banach space 
Question: Let $\mathcal{L}$ be the normed space of all Lipschitz functions on a Banach space $X$ that are equal to $0$ at the origin, under the norm 
          $$\|f\| = \sup\left\{ \frac{|f(x)-f(y)|}{\|x-y\|}:x,y\in X \right\}.$$
          Show that $\mathcal{L}$ is a Banach space. 

My attempt: 
Let $(f_n)_{n=1}^\infty$ be a Cauchy sequence in $(\mathcal{L},\|\cdot\|).$
        Define $f:X\to \mathbb{R}$ by 
        $$ f(x) = \lim_{n\to\infty} f_n(x).
  $$
        We claim that $f$ is well-defined, that is, the limit exists. 
        Fix $x\in X\setminus \{0\}$ and $\epsilon>0.$
        Since $(f_n)_{n=1}^\infty$ is Cauchy, there exists $N\in \mathbb{N}$ such that for any $m,n\geq N,$
$$ \|f_m-f_n\| <\frac{\epsilon}{\|x\|}.
  $$
        Note that 
        $$ \|f_m-f_n\| = \sup_{x,y\in X} \frac{|(f_m-f_n)(x) - (f_m-f_n)(y)  |}{\|x-y\|} \geq \frac{|f_m(x) - f_n(x)|}{\|x\|}.
  $$
        So we have $|f_m(x) - f_n(x)| < \epsilon.$
        Therefore, $(f_n(x))_{n=1}^\infty$ is Cauchy in $\mathbb{R}.$
        Since $\mathbb{R}$ is complete, so $\lim_{n\to\infty} f_n(x)$ exists, that is, $f$ is well-defined.
We claim that $f\in \mathcal{L}.$
        By construction, $f(0) = 0.$
        Since $(f_n)_{n=1}^\infty$ is Cauchy in $(\mathcal{L},\|\cdot\|),$ therefore  it is bounded, that is, there exists $B>0$ such that $\|f_n\|\leq B$ for all $n\geq 1.$
        Since 
$$f = f_N + (f-f_N),$$ 
so for any $x,y\in X,$
\begin{align*}
  |f(x)-f(y)| & \leq |f_N(x)-f_N(y)| + |(f-f_N)(x) - (f-f_N)(y)| \\
  & \leq \|f_N\|\| x-y\| + \|f-f_N\| \|x-y\| \\
  & \leq \|f_N\|\| x-y\| + \|x-y\| \\
  & = (\|f_N\|+1)\|x-y\|.
  \end{align*}
        Therefore, $\|f\|$ is finite and hence $f\in \mathcal{L}.$
Lastly, we wish to prove that $(f_n)$ converges to $f$ in $(\mathcal{L},\|\cdot\|).$
        Indeed, for every $\epsilon>0,$ by Cauchyness of $(f_n),$ there exists $N\in\mathbb{N}$ such that for any $m,n\geq N,$
$$ \|f_n-f_m\|<\frac{\epsilon}{2}.
  $$
        Then 
        $$ \|f-f_n\| = \|\lim_{m\to\infty} f_m - f_n\| = \lim_{m\to\infty} \|f_m-f_n\| < \epsilon.
  $$
        We conclude that $\mathcal{L}$ is complete.

Is my proof above correct?
 A: Your proof has all of the right ideas but suffers from a couple of defects since you end up implicitly doing things like assuming $f \in \mathcal{L}$ before you've proved it and confusing the pointwise limit with the limit in $\mathcal{L}$. To fix this, you just need to write everything down in terms of the definition of the norm and make sure you're working with pointwise limits. I include details below.
Firstly, when trying to prove that $f \in \mathcal{L}$, on the second line you write down $\|f-f_N\|$ and then bound this by $1$. Doing this already requires you to assume $f-f_N \in \mathcal{L}$ and hence $f \in \mathcal{L}$ and also something like $f_n \to f$ in $\mathcal{L}$. 
Instead, write
\begin{align*}
|f(x) - f(y)| &= |\lim_{n \to \infty} (f_n(x) - f_n(y))|
\\&= \lim_{n \to \infty} |f_n(x) - f_n(y)|
\\& \leq \lim_{n \to \infty} \|f_n\| |x-y|
\\& \leq C|x-y|
\end{align*}
where in the last line we used that $\|f_n\|$ is a bounded sequence. Note that in the second line, I only use that $f_n \to f$ pointwise and not in $\mathcal{L}$.
This is important because when trying to prove that $f_n \to f$, you take a pointwise limit and take it outside of the $\mathcal{L}$-norm. This doesn't work since you can only do that once you know that the limit converges in $\mathcal{L}$ and not just pointwise. Instead, work again with a pointwise limit by writing
\begin{align*}
|f(x) - f(y) - f_n(x) + f_n(y)| &= \lim_{m \to \infty} |f_m(x) - f_m(y) - f_n(x) + f_n(y)|\\
& \leq \lim_{m \to \infty} \|f_m - f_n\| |x-y|
\end{align*}
Now since $(f_n)$ is Cauchy in $\mathcal{L}$, there is $N$ independent of $x,y$ such that for $n,m \geq N$ $\|f_n - f_m\| \leq \varepsilon$. So we get
$$|f(x) - f(y) - f_n(x) + f_n(y)| \leq \varepsilon |x-y|$$
which implies for $n \geq N$, $\|f-f_n\| \leq \varepsilon$ so $f_n \to f$ in $\mathcal{L}$.
