Which of the following is a metric on S? d^2 or d^(1/2)

For [tex]d[/tex] to be a metric you need to show that:
[tex]d(a,b) = 0 \iff a=b[/tex]
(which is easy in this case)
[tex]d(a,b) \geq 0[/tex]
(also easy)
[tex]d(a,c) \leq d(a,b)+d(b,c)[/tex]
Which is the only one that really needs any looking into in this case.

The does not necessarily hold for [tex]d'=d^2[/tex] since if you have [tex]d(a,b)=1[/tex] and [tex]d(b,c)=1[/tex] and [tex]d(a,b)=2[/tex], then the triangle inequality does not hold for [tex]d'[/tex].

To prove that the triangle inequality holds for [tex]d'=d^{\frac{1}{2}}[/tex], start with the triangle inequality for [tex]d[/tex], complete the square on the RHS, and take the square root of both sides.