In a Noetherian topological space, a constructible set is a finite union of locally closed sets. This is a conclusion on constructible sets:
Every constructible set contains a dense open subset of its closure.
Now neglect the Noetherian condition, and give $\mathbb{R}$ the usual topology. $\{0\}$ is closed, thus "constructible". The only open set contained in $\{0 \}$ is the empty set. It is certain that the empty set is not dense in the closure of $\{0 \}$ which is equal to $\{0 \}$.
So, the Noetherian condition is necessary.
But the result does not seem clear to me. Would you please give me a proof or a reference?
Many thanks.