topological property If $p$ is a topological property, then a space $(X, \tau)$ is said to
be minimal $p$ (respectively, maximal $p$) if $(X, \tau)$ has property $p$ but no topology
on $X$ which is strictly smaller (respectively, strictly larger) than $\tau$has $p$.
The  spaces are called  $KC$-spaces in which every  compact subset is closed.
The  spaces are called strongly $KC$-spaces in which every countably compact subset is closed.

(1) Is the property of being strongly minimal KC space a topological property?
(2) We know that a minimal KC - space is  countably compact.
So, can we say a closed subspaces of a minimal KC- space is minimal KC? 

.

If $p$ is a topological property, then so is the property of being minimal $p$.
Suppose that $\langle X,\tau\rangle$ be minimal $p$. Let $\langle Y,\sigma\rangle$ be a space homeomorphic to $\langle X,\tau\rangle$, and let $h:X\to Y$ be a homeomorphism. Suppose that $\sigma'$ is a strictly smaller $p$ topology on $Y$. Let $\tau'=\{h^{-1}[U]:U\in\sigma'\}$. Then $\tau'$ is a topology on $X$ strictly smaller than $\tau$, and $h$ is a homeomorphism from $\langle X,\tau'\rangle$ to $\langle Y,\sigma'\rangle$. The property $p$ is topological, so $\langle X,\tau'\rangle$ has $p$, contradicting the assumption that $\langle X,\tau\rangle$ was minimal $p$. Thus, no such topology $\sigma'$ exists, and $\langle Y,\sigma\rangle$ is minimal $p$.
The strong $KC$ property is clearly topological, so the property of being minimal strongly $KC$ is also topological.

Let $\langle X,\tau\rangle$ be minimal $KC$; we actually know that $X$ is compact, by the result of Bella & Costantini that I cited in this answer. Every compact $KC$ space is minimal $KC$, and every closed subset of $X$ is compact and $KC$, so every closed subset of $X$ is minimal $KC$.
