Locally Closed Subset is Closed

Question

I have a question that arises from following former question of mine: Question about Argument from Mumford's Red Book

There I wanted to find out under which conditions for a $k$-variety $X$ a certain locally closed set $Z \subset X \times_{Spek(k)}\ X$ is already closed.

The solution was (still not satisfying) that if we denote by

$$Cl := \{x \in X \times_{Spek(k)} X \vert x \text{ is a closed point in } X \times_{Spek(k)} X \}$$ the set of closed points (which is dense) then $Z$ is closed in $X \times_{Spek(k)} X$ if the intersection $Z \cap Cl$ is closed in $Cl$ with respect to induced subset topology on $Cl$.

Now I have following QUESTION: Is is possible to verify this statement pure topologically? So without any considerations on variety/scheme structure?

Or in other words: Does following statement hold and why?

Let $X$ be a topological space, $L \subset X$ a locally closed subset in $X$ and $D$ a dense subset.

Assume that $L \cap D $ is closed in $D$ wrt subset topology. Is then $L$ already closed in $X$? Why?

Remark: here I work with following definition(s) for locally closed: https://ncatlab.org/nlab/show/locally+closed+set

Answer 1

You ask:

If $L\subset X$ is a locally closed subset in $X$ and $D$ a dense subset. If $L\cap D$ is closed in $D$ with respect to the subspace topology. Is then $L$ already closed?

No. Notice an open set is locally closed. (a proof can be found here or here). Let $L=(\sqrt2,\sqrt3)$ and $D=\Bbb Q$. Then $L\cap D=(\sqrt2,\sqrt3)\cap \Bbb Q=[\sqrt2,\sqrt3]\cap \Bbb Q$, hence closed in $D$. But $(\sqrt2,\sqrt 3)$ is obviously not closed in $\Bbb R$.