Halp

How do I proved this?

Assume h : R -> R is continuous on R. Let K = {x : h(x)=0}. Show that K is a closed set.

>>1
Let L be the compliment of K, so L = {x : h(x) != 0}. Let x0 be a point in L. Then |h(x0)| > e for some positive real number e. By the definition of continuity, there exists a positive real number d such that for all x satisfying |x0 - x| < d, |h(x0) - h(x)| < e. From the above fact that |h(x0)| > e, we have |h(x)| > 0. Hence the open ball of radius d around x0 is in L, and L is open. K is the compliment of an open set, and thus is a closed set.