Find finite and r-dense subset in the unity ball of $C[0,1]$. 
We say that $Y\subset(X,d)$ is $r$-dense if $\forall x\in X\ \exists\ y\in Y:d(x,y)<r$. Find for what values of $r$ there is a finite and $r$-dense subset in the unity ball of $C([0,1])$ with $d_{\infty}$.

What I did:
First, the distance given is $$d((x_1,y_1),(x_2,y_2))=\operatorname{sup}\{d(x_1,x_2),d(y_1,y_2)\}$$ and $$B_{r}(f_0)=\{f:[0,1]\to[0,1]:d_{\infty}(f_0,f)<r\}$$
then
$$B_{r}(f_0)=\{f:[0,1]\to[0,1]:\sup\{|f(x)-f_0(x)|\}<r, \;\forall x\in [0,1]\}$$
Now I believe that I don't understand the problem, because it asks if for a given $r$ there is $Y\subset B_r(f_0)$ for any $f_0\in [0,1]^{[0,1]}$ there is $f$ such that $\operatorname{sup}\{|f(x)-f_0(x)|\}<r$, which would be true for every $r$ since every $f$ inside the ball satisfies the condition given, isn't enough then to take a finite amount of such functions to have the $Y$ desired?.
 A: For simplicity, I will only talk about the ball around $f_0 \equiv 0$.
The problem asks the following: for a given $r>0$, can we find a set $Y = f_1,\dots,f_n$  such that for any continuous $f$ with $\sup_{x \in [0,1]} |f(x)| < 1$, there is a function $f_k \in Y$ such that $\sup_{x \in [0,1]}|f_k(x) - f(x)| < r$?
This does not seem to match up with what you've said.

Towards a solution: 
Trivially, we can set $Y = \{f_0\}$ for $r = 1$.  
It is not, however, possible for any $r < 1$.
Proof of impossibility: fix $r < 1$.  Consider any finite set $Y = \{f_1,\dots,f_n\}$.  Define the points $x_i = 1/i$ for $i = 1,\dots,n$.
For each $i$, we note that $S_i := [-1,1] \setminus [f_i(x)-r,f_i(x)+r] \neq \emptyset$.  Select $y_i \in S_i$.
We now have the points $(x_i,y_i)$ for $i = 1,\dots,n$ each of which satisfies $|f_i(x) - y_i| > r$.  We may interpolate these points linearly to get a new continuous function $f$.
We find that $d_\infty(f,f_i) > r$ for each $i$ from $1$ to $n$, yet $f \in C[0,1]$.
We therefore conclude that $C[0,1]$ has no finite $r$-dense subset if $r < 1$.
