I would like to prove that given $X \subset \mathbb{P}^n$ quasi-projective variety, i.e. a locally closed subset, every open and closed subsets of X are quasi-projective varieties.
Let $U\subset X$ be an open subset, then $U\subset X\subset \overline{X}$, so $U$ is a open set within a closed set, so it is locally closed.
Let $A\subset X$ be a closed subset, then $A\subset X$ and I think that if I can prove that I can always find an open set $V$ such that $A\subset V\subset X$ then $A=A \cap \overline{V}$ is a locally closed set. But I am not able to determine if this is right and how to prove it.
Have you any suggestions?
This is probably not the easiest way to show what you want but here is what I thought about:
For any $x$ in $A$, $x$ is also in $X$ so there is an open $U \in \mathbb{P}^n$ such that $X\cap U$ is closed in $U$ (definition of locally closed subset, eg the equivalency of two definitions of locally closed sets)
Now I claim that $A\cap U$ is closed in $U$ as it is closed in $X\cap U$. Indeed $(X \cap U) \setminus (A\cap U) = (X\setminus A) \cap U$ is open in $X\cap U$ as it is open in $X$ by definition of subset topology.