Why is $L^{\infty}$ not separable? $l^p (1≤p<{\infty})$ and $L^p (1≤p<∞)$ are separable spaces.

What on earth has changed when the value of $p$ turns from a finite number to ${\infty}$?

Our teacher gave us some hints that there exists an uncountable subset such that the distance of any two elements in it is no less than some $\delta>0$. 
Actually I don't understand the question very well, but I hope I have made the question clear enough.
Thank you in advance.
 A: To be separable means to have a countable dense subset.  Suppose that $(M, d)$ is a metric space and that  $U\subseteq M$ be an uncountable subset and $r > 0$. Suppose that for all $x \neq  y \in U$, $d(x,y) \ge r$.  Let $C$ be any countable subset of $M$.  Then $C$ can only meet a countable number of the balls $B_{r/2}(x)$ for $x\in U$.  Let $G$ be the union of all $B_{r/2}(x)$ for $x\in U$ that do not meet $C$.  $G$ is a nonempty open subset of $M$ that does not meet $C$.  
There can be no countable dense subset of $M$.
Consider the case of $\ell^\infty$.  For each subset $Q$ of the integers, let $x_Q$ be the sequence that is 1 on $G$ and 0 off of it.  The $x_Q$ are uncountable and any two elements of this collection are distance 1 apart.  We have just shown that $\ell^\infty$ is not separable.  
You can generate a similar construct for $L^\infty$.  Consider the uncountable subclass of characteristic functions $\{\chi_{B_r(0)}\}_{r>0}\subseteq L^\infty(\mathbb{R}^n)$. Then each pair of distinct elements in it would be 1 unit distance apart. Ergo there cannot be any countable subset of $L^\infty(\mathbb{R}^n)$ that is dense in it.
A: With the topology $\|f-g\|_{l^\infty}=\sup_i |f_i-g_i|$ it is easy to show if C is any countable set, and $(f_{ij})_j$ a sequence of elements of $C$, you can take an element $h\in l^\infty$ so that
$h_i= 0$ if $|f_{ii}|>1$, $h_i=2$ otherwise, hence $\|f_{j}-h\|>1/2$ for all $j$. Then no ball of radius $1/2$ around $h$ can contain an element of $C$.
