If I have an open cover, can I choose a countable sub-cover such that the union of the open sets strictly increasing? If I have an open cover, can I choose a countable  sub-cover such that  the union of the open sets strictly increasing?

Actually I want to use this to prove every noetherian space is compact. Thanks!
In general you can't: Consider $X = [0, 1]$ (which is uncountable) with the topology in which all sets are open. The singletons (= sets of the form $\{x\}$, where $x \in [0, 1]$) form an open cover, but there does not exist a countable subcover.

In the case of Noetherian spaces, we can, in fact, find a finite subcover by failing to find a subcover as you described (in fact, we prove that even in Noetherian spaces it can not exist). Let $X$ be a Noetherian space. Pick an open cover $O_\upsilon, \upsilon \in \Upsilon$ of $X$ and pick $O_1 \in \{O_\upsilon | \upsilon \in \Upsilon\}$ arbitrarily, then $O_2$ such that $O_2 \setminus O_1 \neq \emptyset$, then $O_3$ such that $O_3 \setminus (O_1 \cup O_2) \neq \emptyset$ and so on. Finding such an infinite sequence is equivalent to finding a subcover which you described.

Now given $O_1, \ldots, O_k$, it is possible to find $O_{k+1} \in \{O_\upsilon | \upsilon \in \Upsilon\}$ such that
$$
O_{k+1} \setminus \bigcup_{j=1}^k O_j \neq \emptyset
$$
if and only if the sets $O_1, \ldots, O_k$ do not cover $X$. So if we assume that we never arrive at a $k$ where $O_1, \ldots, O_k$ cover $X$, we can find a sequence $U_n$ of open sets in $X$ defined by
$$
U_n := \bigcup_{j=1}^n O_j
$$
which is strictly increasing. Hence, the sequence of closed sets $A_n$ defined by
$$
A_n := X \setminus U_n
$$
is strictly decreasing. But in Noetherian spaces, there is no strictly decreasing countable chain of closed sets.

Note that we used the axiom of dependent choice for our construction of the sequence $O_n$.
Let $(X,\large\tau),$ be a Noetherian Topology then it is compact

Let $(X,\large\tau)$ be a Noetherian topology  and let $\{U_\alpha\}$ be a cover of $M\subset X$. Set $V_1=U_i$ for any $U_i\in \{U_\alpha\}$ and define $V_{i+1}=V_i\cup U_k$ such that $U_k\in \{U_\alpha\}$ satisfying $V_i\subsetneq V_{i+1}$. Then it follows that $\{V_i^c\}$ is a sequence of closed sets such that $V_{i+1}^c\subsetneq V_i^c$. If $\{V_i^c\}$ is infinite, since $(X,M)$ is a Noetherian topology, it follows that there exist an $N\in Z^+$ such that $i\geq N$ implies $V_i=V_{i+1}$ so $\{V_i^c\}$ is finite. Thus for any open over $\{U_\alpha\}$ of any set $M$ in $(X,\large\tau)$, there can only exist a finite collection of open sets $\{U_{k_j}: U_{k_j}\in \{U_\alpha\}\}$ such that $\{V_i: V_{i}=\bigcup U_{k_j}\}$ is a strictly increasing collection of open sets. Thus letting $M=X$, and we see that $(X,\large\tau)$ is compact.