I will state some definitions to clarify my question.
Definition 1 Let $X$ be a topological space. If every open cover of $X$ has a finite subcover, $X$ is called quasi-compact.
Definition 2 Let $X$ be a topological space. If every non-empty set of open subsets of $X$ has a maximal element, $X$ is called noetherian.
Definition 3 Let $X$ be a topological space. If every point of $X$ has an open neighborhood which is noetherian, $X$ is called locally noetherian.
My question Can we prove that a quasi-compact locally noetherian space is noetherian without Axiom of Choice?
Remark I'm particularly interested in the following question. Let $k$ be a field. Let $X$ be a scheme of finite type over $k$. Can we prove that the underlying topological space of $X$ is noetherian without Axiom of Choice?
If the title questionj is affirmative, the question is also affirmative. See the David Speyer's answer to this question.
As for why I think this question is interesting, please see(particularly Pete Clark's answer): Why worry about the axiom of choice?