Cartesian product of KC spaces We know that Cartesian product of KC spaces do not need to be a KC space.
Is it true to say" if $X$ is a $KC$-space then for each $k ‎\geq2$ ,$X^k$ is $KC$ iff each compact subspace of $X$ is Hausdorff? Why?
 A: Half of the result follows from this generalization of the argument that I made in this answer.

Proposition. If $X$ is a compact $KC$ space that is not Hausdorff, then $X\times X$ is not $KC$.
Proof. Suppose that $\langle X,\tau\rangle$ is a compact $KC$-space that is not Hausdorff, and let $p$ and $q$ be two points of $X$ that cannot be separated by disjoint open sets. Let $\Delta=\{\langle x,x\rangle:x\in X\}$, the diagonal in $X\times X$; $\Delta$ is homeomorphic to $X$, so $\Delta$ is compact; I’ll show that $\Delta$ is not closed and hence that $X\times X$ is not $KC$.
Let $z=\langle q,p\rangle\in(X\times X)\setminus\Delta$, and let $U$ be any open nbhd of $z$ in $X\times X$; there are $V_p,V_q\in\tau$ such that $p\in V_p$, $q\in V_q$, and $V_p\times V_q\subseteq U$. By the choice of $p$ and $q$ we know that $V_p\cap V_q\ne\varnothing$, so let $x\in V_p\cap V_q$; clearly $\langle x,x\rangle\in(V_p\times V_q)\cap\Delta\subseteq U\cap\Delta$, so $z\in(\operatorname{cl}\Delta)\setminus\Delta$. Thus, $\Delta$ is a compact subset of $X\times X$ that is not closed, and $X\times X$ is not $KC$. $\dashv$
Corollary. If the $KC$ space $X$ has a non-Hausdorff compact subset $K$, then $X\times X$ is not $KC$.
Proof. $K\times K$ is a subspace of $X\times X$, and the $KC$ property is hereditary. $\dashv$.

Now suppose that every compact subset of $X$ is Hausdorff, and let $K$ be a compact subset of $X^k$ for some finite $k\ge 2$. Let $K_i$ for $i=1,\dots,k$ be the projections of $K$ to the factors; then $K\subseteq\prod_{i=1}^kK_i$, and $\prod_{i=1}^kK_i$ is compact and Hausdorff. The compact set $K$ is therefore closed in $\prod_{i=1}^kK_i$. $X$ is $KC$, so each $K_i$ is closed in $X$, and $\prod_{i=1}^kK_i$ is closed in $X^k$. Thus, $K$ is a closed subset of a closed subset of $X^k$ and is therefore closed in $X^k$, which is therefore $KC$.
