As Chris Janjigian said, the passage you quoted is a proof that Lipschitz functions are dense; perhaps one should say at the end
Thus, if $X$ is compact, then $\operatorname{Lip}X$ is dense in $C(X)$ by the Stone-Weierstrass theorem.
But to directly answer the question posed in the title: a constructive self-contained proof (without Stone-Weierstrass) goes as follows. Given a continuous function $g$ and a number $\epsilon>0$, pick $\delta>0$ such that $|g(x)-g(y)|<\epsilon$ whenever $d(x,y)<\delta$ — this is possible by uniform continuity of $g$. Also let $M=\sup_X |g|$. Define
$$f(x) = \sup_{y\in X} \left(g(y)- 2M \delta^{-1}d(x,y)\right)\tag{1}$$
I claim that $f$ is Lipschitz and $\sup_X |f-g|\le \epsilon$.
- $f(x_1)-f(x_2)\le 2M \delta^{-1}d(x_1,x_2)$ by the triangle inequality. Reversing the roles of $x_1,x_2$, we see that $f$ is Lipschitz with constant $2M \delta^{-1}$.
- We have $f(x)\ge g(x)$, because the expression under the supremum in (1) turns to $f(x)$ when $y=x$.
- If $d(x,y)\ge \delta$, then the expression under the supremum in (1) is at most $-M$, which is less than $g(x)$.
- If $d(x,y)< \delta$, then the expression under the supremum in (1) is at most $g(x)+\epsilon$, by the choice of $\delta$.
- Combine the items 2-3-4 to obtain $g(x)\le f(x)\le g(x)+\epsilon$.
Let $(f_n)\subseteq\operatorname{Lip}(X,d)$ be a Cauchy sequence. In particular, $(f_n)$ is a Cauchy sequence of continuous functions with respect to the sup norm, so it converges uniformly to a (continuous) function $f$ (this is what you already knew). Let's show that $f_n\rightarrow f$ in $\operatorname{Lip}(X,d)$.
Given $\varepsilon>0$, choose $N\in\mathbb{N}$ such that $\Vert f_n-f_m\Vert<\varepsilon$ whenever $n,m\geq N$. Given $n,m\geq N$ and $x\neq y$ in $X$, we have
$$\frac{|f_n(x)-f_m(x)-(f_n(y)-f_m(y))|}{d(x,y)}\leq \Vert f_n-f_m\Vert_d\leq \Vert f_n-f_m\Vert<\varepsilon.$$
Letting $m\rightarrow \infty$ and taking the sup on $x\neq y$, we obtain that $\Vert f_n-f\Vert_d<\varepsilon<\infty$ for any $n\geq N$, so (given one such $n$), we have $f_n-f\in\operatorname{Lip}(X,d)$, thus $f=f_n-(f_n-f)\in\operatorname{Lip}(X,d)$. Furthermore, we have just proved that $\Vert f_n-f\Vert_d\rightarrow 0$ as $n\rightarrow\infty$, and we already knew that $\Vert f_n-f\Vert_\infty\rightarrow 0$ as $n\rightarrow\infty$. This means exactly that $f_n\rightarrow f$ in $\operatorname{Lip}(X,d)$.
Therefore, $\operatorname{Lip}(X,d)$ is a Banach space.
Now, let $f,g:X\rightarrow\mathbb{R}$. Let $x\neq y$ in $X$. Then
$$|(fg)(x)-(fg)(y)|\leq|f(x)||g(x)-g(y)|+|f(x)-f(y)||g(y)|$$
$$\leq\Vert f\Vert_{\infty}\Vert g\Vert_d d(x,y)+\Vert f\Vert_d d(x,y)\Vert g\Vert_\infty.$$
So given any $z\in X$,
$$|(fg)(z)|+\frac{|(fg)(x)-(fg)(y)|}{d(x,y)}\leq\Vert f\Vert_\infty\Vert g\Vert_\infty+\Vert f\Vert_\infty\Vert g\Vert_d+\Vert f\Vert_d\Vert g\Vert_\infty$$
$$\leq (\Vert f\Vert_\infty+\Vert f\Vert_d)(\Vert g\Vert_\infty+\Vert g\Vert_d).$$
Taking the sup on $x\neq y$ and $z\in X$, we obtain $\Vert fg\Vert\leq\Vert f\Vert\cdot\Vert g\Vert$.
Remark: Maybe this exercise can be easier if you first show that $\Vert f\Vert_d=\inf\left\{K\geq 0:\forall x,y\in X,\ |f(x)-f(y)|\leq Kd(x,y)\right\}$, and then use that expression.
Best Answer
Any compact set in a metric space is separable. Hence Lip1 is separable and so is its closure.
If $D_n$ is a coutable dense set in $\{f\in Lip1:\|f\|_{Lip1} \leq n\}$ then $\cup_n D_n$ is a countable dense set in $C[0,1]$.