Zariski space which never occurs as the underlying topological space of a scheme. According to Hartshorne's "Algebraic Geometry", we say that a topological space $X$ is a Zariski space if it is noetherian (i.e. $X$ satisfies the descending chain condition for closed subsets) and every nonempty closed irreducible subset has a unique generic point (Exercise II.3.17.). In this exercise, we show that the underlying topological space of a noetherian scheme is a Zariski space and here arises a question: conversely, every Zariski space occurs as an underlying topological space of some scheme?
I know that if we remove the Zariski assumption and require only to be noetherian, this assertion does not hold, because, for example, $X=\{x,y\}$ with indiscrete topology gives an counterexample. Indeed, this space is obviously noetherian but never appears as an underlying space of scheme because $X$ has no closed points. However, when $X$ is a Zariski space, I can neither find a counterexample nor prove this statement.
Thank you!

Every Zariski space is the underlying topological space of an affine scheme.  This follows from Hochster's classification of spectral spaces (spaces that are homeomorphic to Spec of a ring) as exactly the sober spaces in which compact open sets are closed under finite intersections and are a basis for the topology.  In a Noetherian space, every subset is compact, so every Noetherian sober space is a spectral space.
Note, though, that not every Noetherian sober space is the underlying topological space of a Noetherian scheme.  For instance, a Noetherian ring of dimension greater than $1$ always has infinitely many prime ideals (this is a consequence of Krull's principal ideal theorem and prime avoidance; see here for instance).  So, a finite sober (or equivalently, just finite $T_0$) space of dimension greater than $1$ cannot be the underlying topological space of a Noetherian scheme, even though it is the spectrum of some (non-Noetherian!) ring.
