Atiyah-Macdonald Ex7.22 Let $X$ be a Notherian topological space and let $E$ be a subset of $X$. Show that $E$ is open in $X$ if and only if, for each irreducible closed subset $X_0$ in $X$, either $E \cap X_0 = \emptyset$ or else $ E \cap X_0 $ contains a non-empty open subset of $X_0$.
$\Rightarrow$ direction is easy and I'm trying to solve $\Leftarrow$ direction by contrapositve. So assume that $E$ is not open in $X$. To use the Noetherian condition, I came up with the set of closed sets X' \subseteq X such that E\cap X' is not open in $X$. Then I showed that the minimal element $X_0$ of this set is irreducible, and $E\cap X_0 \neq \emptyset$. But there seems no way to deduce a contradiction if $E\cap X_0$ contains a non-empty open subset of $X_0$.
So I come up with the set of closed sets X' \subseteq X such that E\cap X' is not open in X'. Then the others can be proved, but now I can't show that $X_0$ is irreducible. If $X_0=C_1 \cup C_2$ with proper closed subsets of $X_0$ then $E\cap X_0 = (E \cap C_1) \cup (E \cap C_2)$. By the minimality of $X_0$, $E \cap C_i$ is open in $C_i$ so that $E \cap C_i=U_i \cap C_i$. But I think it needs not be true that $(U_1 \cap C_1) \cup (U_2 \cap C_2)$ is open in $X_0$, by sketching some sets in $\mathbb{R}^2$.
So how can I solve the problem? Or is there any way to correct my above trials?