Well, for your set to be a subspace, it has to be closed under addition and scalar multiplication. For the first, this means that for f(x) and g(x) in the set, f(x)+g(x) must also be in the set. For the second, for f(x) in the set, then c*f(x) must also be in the set for any scalar c.
A side effect of this second condition is that the zero function, f(x)=0 MUST be in the set for it to be a subspace in R.
Now, the zero condition is easy to prove, and so are the others. So try it, I suppose.
EDIT: As an example of a proof, take f(x)+g(x). Well, for the function to be in the subspace, we know that (f(1)+g(1))(f(2)+g(2)) must be zero. Now just as an example of functions that are in the subspace, let's take f(x)=1-x and g(x)=2-x. We can see that f(1)=0, f(2)=-1, g(1)=1, and g(2)=0. Thus, we need (0+1)(-1+0) to be zero, but it's not, it's -1.
So that's a rather effective counterexample.