Can we prove that a quasi-compact locally noetherian space is noetherian without Axiom of Choice? 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?
 A: Yes, we can.
Consider $\mathcal U$ as the open cover whose elements are all the open neighborhoods witnessing the local noetherian property. Since every $x\in X$ has such neighborhood, this is certainly an open cover. By quasi-compactness, we have some finite subcover, $V_1,\ldots,V_n$.
Now let $\mathcal V$ be any non-empty family of open sets. Write $\mathcal V_i=\{U\cap V_i\mid U\in\mathcal V\}$. Then there is at least one $i$ such that $\mathcal V_i$ is not empty, and by the noetherian property of $V_i$ there is a maximal open set in $\mathcal V_i$. Of course, there might be more than just $j$ such that $\mathcal V_j$ is non-empty.
For each $U\in\mathcal V$ write $m(U)=|\{i\mid U\cap V_i\text{ is maximal}\}|$, that is $m(U)$ is the number of $i$'s in which $U\cap V_i$ is maximal in $\mathcal V_i$. Since the values of $m(U)$ are between $0$ and $n$ there is some maximal value $m$, and it has to be at least $1$.
Finally, consider $\{U\in\mathcal V\mid m(U)=m\}$. Somewhere in this set we have a maximal open set in $\mathcal V$. Pick some $U_0$, if it is maximal then we are done. Otherwise, consider $\mathcal U_0=\{V\in\mathcal V\mid U_0\subseteq V\}$, and note that for each such $V$ we have $m(V)=m(U_0)=m$, and in particular where $U_0\cap V_i$ is maximal, $V\cap V_i=U_0\cap V_i$.
Let $i$ be the least index such that $U\cap V_i$ is not maximal, and there is some $V$ such that $U\cap V_i\subsetneq V\cap V_i$ (such index exists, otherwise $U$ is maximal), now consider $\{V\cap V_i\mid V\in\mathcal U_0\}$ and let $\mathcal U_1$ be the set of $V$ such that $V\cap V_i$ is maximal. Pick some $U_1\in\mathcal U_1$, if it is maximal, then we are done.
Otherwise we proceed to repeat the process on the least index $i$ such that for some $V\in\mathcal U_1$ we have $U_1\cap V_i\subsetneq V\cap V_i$. We observe that $U_0\subseteq U_1$, and the induction process guarantees that we have an increasing chain of open sets.
However the induction can only proceed $n-m$ steps, so it must halt and we must have a maximal element.
A: Since $X$ is quasi-compact and locally noetherian, there exists a finite open cover $V_1,\ldots,V_n$ each of which is noetherian.
Let $\mathcal V$ be any non-empy family of open sets.
Write $\mathcal V_i=\{U\cap V_i\mid U\in\mathcal V\}$.
We define inductively an element $U_i \in \mathcal V_i$ and a non-empty subset $\mathcal U_i$ of $\mathcal V$ for each $i$.
Since $V_1$ is noetherian, there exists a maximal element $U_1$ of $\mathcal V_1$.
Let $\mathcal U_1 = \{U \in \mathcal V\mid U\cap V_1 = U_1\}$.
Let $i$ be an integer such that $1 \lt i \le n$.
Suppose $U_j$ and $\mathcal U_j$ are defined for all $j$ such that  $1 \le j \lt i$
Let $U_i$ be a maximal element of $\{U\cap V_i\mid U\in\mathcal U_{i-1}\}$.
Let $\mathcal U_i = \{U \in \mathcal U_{i-1}\mid U\cap V_i = U_i\}$.
Notice that $\mathcal U_i$ is not empty since $V_i$ is noetherian.
Now that $U_i$ and $\mathcal U_i$ are defined for all $i(1 \le i \le n)$, it suffices to prove that any element $W$ of $\mathcal U_n$ is a maximal element of $\mathcal V$.
Let $V$ be an element of $\mathcal V$ such that $W \subset V$.
It suffices to prove that $W = V$.
We first prove by induction that $V \in \mathcal U_i$ for all $i = 1, \cdots n$.
Since $\mathcal U_n \subset \mathcal U_{n-1} \subset \cdots \subset \mathcal U_1 \subset \mathcal V$, $W \in \mathcal U_i$ for all $i = 1, \cdots, n$.
Since $W\in \mathcal U_1$, $W\cap V_1 = U_1$.
Since $U_1$ is a maximal element of $\mathcal V_1$ and $V\in \mathcal V$,
$W\cap V_1 = V\cap V_1 = U_1$.
Hence $V\in \mathcal U_1$.
Let $i$ be an integer such that $1 \lt i \le n$.
Suppose $V \in \mathcal U_j$ for all $j$ such that $1 \le j \lt i$.
Since $W \in \mathcal U_i, W\cap V_i = U_i$.
Since $U_i$ is a maximal element of $\{U\cap V_i\mid U\in\mathcal U_{i-1}\}$ and $V \in \mathcal U_{i-1}$, $V\cap V_i = U_i$.
Hence $V \in \mathcal U_i$
Therefore $W\cap V_i = V \cap V_i = U_i$ for each $i$.
Hence $W = V$ as desired.
QED
