Closed sets in Noetherian topology Let $X$ has a Noetherian topology. Then how can it be proven that $Y \subseteq X$ is closed in $X$ iff there are open sets $\{ W_{\alpha} \}_{\alpha \in A}$ such that $\bigcup_{\alpha \in A}W_\alpha = X$ and $W_{\alpha} \cap Y$ is closed in $W_{\alpha}$ (with respect to the induced topology)?
This is my aproach:
I know that since the topology is Noetherian there are finite number $\{ W_1, W_2, \dots W_n \} \subseteq \{ W_{\alpha} \}_{\alpha \in A}$ such that $W_1 \cup W_2 \cup \dots \cup W_n = X$. So it is sufficient to deal with the case when the set of indeces $A$ is finite. The simplest case when $X = W_1 \cup W_2$ and $W_1 \cap W_2 = \emptyset$ is trivial, but I have difficulty to prove the assumption when $W_1 \cap W_2 \neq \emptyset$. 
Any help with this? Thanks in advance!
 A: If $Y$ is closed, then every $W_\alpha \cap Y$ is also closed, by definition of subspace topology.
If $Y$ is not closed, then we have $Y' = \bar Y \setminus Y \neq \varnothing$, where $\bar Y$ is the closure of $Y$. This $Y'$ must intersect $W_\beta$ for some $\beta$ (since the $W_\alpha$ cover $X$). We then have that the closure of $W_\beta \cap Y$ in $W_\beta$ is equal to $W_\beta \cap \bar Y$ (again by definition of subspace topology, and the definition of closure of $Y$ as the intersection of all closed sets that contain $Y$). Since $Y'\cap W_\beta \neq \varnothing$, we have $W_\beta \cap \bar Y \neq W_\beta \cap Y$, and hence $W_\beta \cap Y$ is not closed in $W_\beta$.

Additional proof that $Z$, the closure of $W_\beta\cap Y$ in $W_\beta$, is equal to $W_\beta\cap \overline Y$:
We have $Z \subseteq W_\beta\cap \overline Y$ trivially, since the latter is a closed subset of $W_\beta$ that contains $W_\beta \cap Y$. It remains to show inclusion the other way.
Now, take any closed set $V_W\subseteq W_\beta$ that contains $W_\beta\cap Y$. This $V_W$ is the intersection of $W_\beta$ with some closed set $V_X \subseteq X$. Since $W_\beta$ is open in $X$, its complement is closed, which means we may assume that $V_X$ contains the entire complement of $W_\beta$.
This specifically means that any such $V_X$ contains all of $Y$. Therefore $\overline Y$ is contained in the intersection of all the $V_X$. Intersecting this statement with $W_\beta$, it says that $W_\beta\cap \overline Y$ is contained in the intersection of al $V_W$, which by definition is $Z$, and we are done.
A: HINT: Suppose that $\{W_1,\ldots,W_n\}$ is a finite open cover of $X$ such that $W_k\cap Y$ is closed in $W_k$ for $k=1,\ldots,n$. If $Y$ is not closed in $X$, there is some point $x\in(\operatorname{cl}_XY)\setminus Y$. Choose $k$ such that $x\in W_k$. $Y\cap W_k$ is closed in $W_k$, and $x\notin Y\cap W_k$, so there is a relatively open $U\subseteq W_k$ such that $x\in U$ and $U\cap Y=\varnothing$. Do you see the contradiction?
