When is a metric space of bounded real valued functions separable? The metric space of bounded, real value functions is $B(S)$, whose domain is $S$.
The metric is $d(f,g) = \sup[|f(s)-g(s)|] : s \in S $. The problem is to show that $B(S)$ is separable iff $S$ is finite.
The authors, Gamelin and Greene of "Introducton to Topology", offer a hint:
For any subsets $T$ of $S$, the balls $B(\mathcal{X}_T,\frac1 2)$ are disjoint, where $\mathcal{X}_T$ is the characteristic function for T. 
I see that if the subsets T are disjoint, then the distances between them are $1$, greater than $1/2$, and consequently the balls are disjoint.   I should like to show that the space $B(S)$ is totally bounded iff S is finite, for then it is separable.   I must then show that any element of $B$ belongs to a finite covering by open balls of $B(S)$ of given radius.  I have not been able to do this with a covering by the balls of the preceding paragraph.
 A: I think first that when $T_1$ and $T_2$ are  distinct subsets of $S$, then the balls $B(1_{T_1},1/2)$ and $B(1_{T_2},1/2)$ are disjoint. In fact, if you have $x\in T_1$ and not in $T_2$, you get $1_{T_1}(x)-1_{T_2}(x)=1$, hence $\|1_{T_1}-1_{T_2}\|\geq 1$ (in fact, it is equal to $1$). Now suppose that you have a subset $A$ of $B(S)$, that is countable and dense. As for each subset $T$ of $S$, there exists an $a$ in $A$ and in the ball of center $1_A$ and radius $1/2$.  Any $b$ in $B(S)-A$ is at least $1/2$ distant and not a limit point of $A$.  Thus, the functions of $B(S)$ are at most countable. But every subset of S may be a domain for a bounded function in $B(S)$.  This shows that the set of all subsets of $S$ is at most countable. Now if $S$ is infinite, there exists a countable subset of $S$, say $C$. Then the set of all the subsets of $C$ is not countable, and this is also the case for the set of all the subsets of $S$: a contradiction! Then $S$ has to be finite. 
A: What do you know if you take $T_1, T_2 \subseteq S$ and take the distance of their characteristic functions? What happens now if $S$ is not finite?

If its finite its trivially separable as you have a finite dense subset, just pick all the functions. So we need the other direction, take $T_1, T_2 \subseteq S$ and $T_1 \neq T_2$ then $d(\mathcal{X}_{T_1}, \mathcal{X}_{T_2}) =1$, now if $S$ is not finite, we have at least an uncountable amount of different characteristic functions and these are all at more than 1/2 of distance, hence there isn't an countable dense subset, since you would have to choose all the functions. 

