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!

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?
