$\int_0^1|f(x)-g(x)|dx=1$ for distinct $f,g\in S$ 
Does there exists an infinite subset $S$ of $C([0,1],\mathbb{R})$ such
  that $$\int_0^1|f(x)-g(x)|dx=1$$ for any distinct $f,g\in S$?

I was guessing the the answer is yes. I can construct such a set with 3 functions, but can't really be generalized.

Let 
$$\triangle_{a,b}=
\begin{cases}
2\frac{x-a}{b-a}, &\text{if }a<x<\frac{a+b}2\\
2\frac{x-b}{a-b}, &\text{if }\frac{a+b}2\le x<b\\
0, &\text{otherwise.}
\end{cases}
$$
The function $\triangle_{a,b}$ is non-negative, zero outside the interval $(a,b)$ and the graph is a triangle of height 1, the integral of this function is the area of the triangle, i.e. $\frac{b-a}2$. (It is good to draw a picture.)
Now you can choose
$$f_n=2^{n+1}\triangle_{1/2^n,1/2^{n+1}}.$$
You get $\int f_n(x)\; \mathrm{d}x=\frac12$ and 
$$\int |f_n(x)-f_m(x)|\; \mathrm{d}x=\int f_n(x)\; \mathrm{d}x+\int f_m(x)\; \mathrm{d}x=1$$ 
for $n\ne m$. (Notice that the support of $f_n$ and the support of $f_m$ are disjoint.)
