Let $X$ be a quasi-compact (not necessarily separated) space and $F\subset X$ a closed subset. I know (hope not mistakingly) that F is quasi-compact.
Conversly, If $F$ is quasi-compact, do we need the separation 2 axiom over $X$ for $F$ to be closed?
Thank you for your help/hint.
Indeed, always we have $X$ (quasi)compact and $F \subseteq X$ closed then $F$ (quasi)compact. Essentially because we can add $X\setminus F$ as an open set to any open cover of $F$ to use the quasi-compactness of $X$ to show that of $F$.
If $X$ is $T_2$ (Hausdorff, or separated, if you prefer) and $F \subseteq X$ is quasi-compact then $F$ is closed in $X$ (no (quasi-)compactness of $X$ is needed) and this fails even for $T_1$ spaces: $X=\mathbb{N}$ in the cofinite topology and $F=\{2n: n \in \mathbb{N}\}$ being a standard example to see this (all subsets of $X$ are quasicompact and only $X$ and the finite subsets are closed).