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Is it true that a dense constructible set of a topological space is open? (It is likely that some conditions are needed on the topological space, maybe noetherianity). How could you prove it?

Edit: a constructible set is a finite union of locally closed sets, and a locally closed subset is the intersection of a closed subset and an open subset.

As is explained here, I know that a constructible set contains a dense open subset of its closure (when the topological space is noetherian), but my question is different.

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You will need some separation axiom, otherwise the result isn't true. Take any topological space $X$ (possibly Noetherian), and let $Y$ be the disjoint union of $X$ and a one-point set $\{*\}$. Consider on $Y$ the topology whose open sets are $\emptyset$ and all sets $U \cup \{*\}$ with $U$ open in $X$. Then EVERY non-empty open set in $Y$ is dense. Now let $U$ be a non-empty open set in $X$, let $C$ be a non-empty closed set in $X$, and choose them so that $U \ \cup C$ is not open in $X$. (In most spaces $X$ you will be able to find such $U$ and $C$).

Now $V = U \cup \{*\}$ is open (and dense) in $Y$, $C$ is closed in $Y$, $V \cup C$ is dense in $Y$, but it is not open (because $U \cup C$ is not open in $X$).

EDIT: I came up with this counter-example while Noah was writing up his, but he posted first. :-)

Take $X = \Bbb R^2$, $U$ the complement of the horizontal axis, and $C$ the origin (or any nonempty, closed, proper subset of the horizontal axis). $U \cup C$ is the counterexample.

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    +1. This is the "right" example, since the whole point is that we can take a constructible dense set and add a point to it (assuming some separation) and this preserves density and constructibility, but generally kills openness. The one I gave doesn't generalize nearly as well. I'll leave mine up for the historical record, though (and for the Noetherian counterexample). – Noah Schweber Aug 05 '16 at 03:47
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It's not true even with separation axioms!

Working in $\mathbb{R}$ with the usual topology, let $X=\mathbb{R}\setminus\{2^{-n}: n\in\mathbb{N}\}$. Clearly $X$ is dense and not open (consider $0\in X$); however, the positive part of $X$ is the intersection of $[0, \infty)$ (a closed set) with $\mathbb{R}\setminus(\{2^{-n}: n\in\mathbb{N}\}\cup\{0\})$ (an open set), and the nonpositive part is just $(-\infty, 0]$ (a closed set); so $X$ is a finite union of locally closed sets, that is, constructible.

Here's a Noetherian counterexample: let $\mathbb{N}$ have the topology where open = coinitial (or empty) - that is, the (nonempty) open sets are exactly the intervals $[n, \infty)$. Then consider $X=\{0\}\cup [7,\infty)$. Clearly $X$ is dense, and not open. But $\{0\}$ is closed, and $[7, \infty)=\mathbb{N}\cap [7, \infty)$ is the intersection of a closed set and an open set, so $X$ is constructible.

Noah Schweber
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    ...and in the second example, the space is Noetherian AND T1 (although not Hausdorff). –  Aug 05 '16 at 15:29