since {0} is a subset of every ring R (since every ring R has an additive identity, called its 0), we need only show the following 2 things (since the set {x in R: x = 0} is non-empty):

1. for all x,y in S = {0}, x - y is in S

2. for all x,y in S, xy is in S

(we automatically get commutativity of +, associativity of +, associativity of *, and distributivity of * over + (both ways) because the operations + and * in S are just those of R, and S is a subset of R).

so:

0 - 0 = 0, which is in S = {0}

0*0 = 0, which is in S = {0}.

that is the entire proof.