 Solution

An affine variety is irreducible when it is nonempty and cannot be expressed as the union of two proper Zariski-closed sets. If a nonempty open subset $U$ were not dense, $X=\overline{U}\cup (X\setminus U)$ would be such a union: the first set is proper by assumption, and the second is proper because $U$ is nonempty. Hence every nonempty open subset of an irreducible variety is dense.