Spec R is not irreducible iff we can find two proper closed subspaces that cover all of spec R. this happens iff the ideals A and B are not in rad(0), so that vA and vB are proper, yet every prime ideal contains either A or B, contains A ∩B, and contains AB. In other words, rad(0) contains AB, but neither A nor B.
In summary, spec R is irreducible iff rad(0) is prime.
Is the closed subspace vE irreducible, for a given ideal E? Again, this space is not irreducible iff we can find vA and vB that cover vE. We may as well assume A and B contain E, since our subspace is restricted to those prime ideals that contain E. Neither A nor B lies in rad(E), else one of our closed sets is the entire space. At the same time, AB lies in rad(E), else vA and vB would miss a prime ideal containing E. Once again rad(E) is not prime.
In summary, vE is irreducible iff rad(E) is prime.
An irreducible closed set is called an irreducible component if it is maximal, i.e. at the top of its chain. This happens iff the corresponding ideal is a minimal prime. The correspondence is one to one. Since every closed set lies in an irreducible component, every radical ideal descends to a minimal prime ideal. (You can see a more direct proof of this fact here.)