$V$ is Noetherian space if only if every open subset of $V$ is compact. 
Let $V$ be a topological space. Show that $V$ is Noetherian space if only if every open subset of $V$ is compact.

$(\Longrightarrow)$ If $W \subset V$ ($W$ open) and $C = \lbrace C_{\alpha} \rbrace$ a cover of $W$, let $F$ be a collection of all finite unions of elements of $C$. Since $V$ is Noetherian, $F$ has a maximal element $C_{\alpha_{1}}\cup...\cup C_{\alpha_{n}}$. If $w \in W\setminus (C_{\alpha_{1}}\cup...\cup C_{\alpha_{n}})$. $C_{\alpha_{1}}\cup...\cup C_{\alpha_{n}} \subsetneq C_{\alpha_{1}}\cup...\cup C_{\alpha_{n}}\cup C_{\beta}$ with $w \in C_{\beta}$, but $C_{\alpha_{1}}\cup...\cup C_{\alpha_{n}}$ is a maximal. Therefore, $W \subset C_{\alpha_{1}}\cup...\cup C_{\alpha_{n}}$ so, $W$ is compact. (Is there an error? Or suggestion?)
$(\Longleftarrow)$ I couldn't prove. Any hint?
 A: If $X$ is Noetherian, let $O\subseteq X$ be open. Let $\mathcal{U}$ be an open cover of $O$. Note that its members (being open in $O$) are also open in $X$.
If $\mathcal{U}$ were an infinite cover, pick $U_1, U_2, \ldots\in \mathcal{U}$ all distinct and non-empty, and then define $V_n = \cup_i^n U_i$ for all $n$ and note that this then would be an infinite chain of open subsets of $X$. This cannot exist, so $\mathcal{U}$ is a finite cover already, and so has itself as a finite subcover. So $O$ is compact.
For the reverse direction: if $U_1 \subseteq U_2 \subseteq U_3 \ldots$ is a chain of open subsets of $X$, then define $O = \cup_{i=1}^\infty U_i$ is open and $\{U_i: i=1,2, \ldots\}$ is by definition an open cover of $O$ which by assumption has a finite subcover $\{U_{i_1}, U_{1_2}, \ldots , U_{i_N}\}$. But then $O = U_j$ where $j = \max(i_1, i_2, \dots, i_N)$ by the $U_i$ being a chain, and so the chain stops at $i_N$.
A: Suppose one has an ascending chain $(U_n)$ of open subsets. The
union $V$ of the $U_n$ is open. As $V$ is compact, and the $U_n$
form an open covering, then finitely many of the $U_n$ cover $V$.
But $(U_n)$ is a chain, so $V=U_n$ for some $n$.
