Let B(S) Be the Described Separable Metric Space of Bounded Real-Valued Functions on the Set S. Is S Finite? If the separable metric space of the title has the metric of uniform convergence on S, defined below, is S finite?  This is expressed as the necessary consequence of the conditional clause in Ex 7 of Chapter 1, Section 5, in the "Introduction to Topology" by Gamelin and Greene, Second Edition, p 25 (Dover, 1999).
The "metric of uniform convergence" is given as: d(f,g)=sup{|f(s)-g(s)|, where s belongs to S  (p.3 Ibid.)
The authors kindly provided a hint, but I have not found it helpful, regrettably.  I have followed the hint to show that the balls of radius 1/2 about the characteristic functions for subsets of S are disjoint, but I have not been able to apply that fact to demonstrate the consequence that S is finite.   
 A: There's a classic way to prove a space is non-separable: construct an uncountable set of disjoint open sets.
How does it work? If there were a countable dense subset of the space, it must intersect with every open set. Since the sets we've constructed are disjoint, each of our open sets must contain a different element of the countable dense subset. But then, our countable dense subset has an uncountable subset, which is a contradiction.
It sounds like you've got most of the rest of the proof.
A: Consider $B(S)$ where $S$ is an infinite set. 
Suppose $D$ is a countable dense subset of $B(S)$ (which means that every open ball contains a point of $D$).
Indeed, define $f_A$ as $f_A(x) = 1$ if $x \in A$, and $f_A(x) = 0$ otherwise.
If $A \neq B$ then $d(f_A, f_B) = 1$, as is easily checked.
This indeed means that $B(f_A, \frac{1}{2})$ , where $A$ ranges over all subsets of $S$, is a disjoint family of open balls. But each of them must contain a (necessarily different) element of $D$ as $D$ is dense.
So $|D| \ge 2^{|D|} >= \aleph_1$ contradiction.
