A topological space is maximal (countably) compact if and WE know that "maximal compact iff $KC$ compact."
Can we say:

A topological space is maximal (countably) compact if and
  only if its (countably) compact subsets are precisely the closed sets. Why?

 A: If $\langle X,\tau\rangle$ is countably compact, then every closed subset of $X$ is countably compact. Thus, we want to show that $\tau$ is a maximal countably compact topology on $X$ iff every countably compact subset of $X$ is closed.
Suppose first that $K$ is a countably compact subset of $X$ that is not closed. Let $G=X\setminus K$, and let $\tau'$ be the topology generated by the subbase $\tau\cup\{G\}$: $$\tau'=\{V\cup(W\cap G):V,W\in\tau\}\;.$$
Let $\mathscr{U}=\{U_n:n\in\omega\}\subseteq\tau'$ be a countable $\tau'$-open cover of $X$. For each $n\in\omega$ let $U_n=V_n\cup(W_n\cap G)$, where $V_n,W_n\in\tau$. Then
$$\begin{align*}
X&=\bigcup_{n\in\omega}U_n\\
&=\bigcup_{n\in\omega}V_n\cup\bigcup_{n\in\omega}(W_n\cap G)\\
&=\bigcup_{n\in\omega}V_n\cup\left(G\cap\bigcup_{n\in\omega}W_n\right)\;,
\end{align*}$$
so $$\bigcup_{n\in\omega}V_n\supseteq X\setminus\left(G\cap\bigcup_{n\in\omega}W_n\right)\supseteq X\setminus G=K\;.$$
$\{V_n:n\in\omega\}$ is therefore a countable $\tau$-open cover of $K$, and $K$ is countably compact in $\langle X,\tau\rangle$, so there is a finite $F_0\subseteq\omega$ such that $$K\subseteq\bigcup_{n\in F_0}V_n\subseteq\bigcup_{n\in F_0}U_n\;.$$ 
Let $H=X\setminus\bigcup_{n\in F_0}V_n$; $H$ is closed and therefore countably compact in $\langle X,\tau\rangle$, and $H\subseteq X\setminus K=G$, so $H\cap W_n\subseteq G\cap W_n$ for each $n\in\omega$.
Now $\{V_n:n\in\omega\}\cup\{W_n:n\in\omega\}$ is a $\tau$-open cover of $H$, so there are finite $F_1,F_2\subseteq\omega$ such that $$H\subseteq\bigcup_{n\in F_1}V_n\cup\bigcup_{n\in F_2}W_n\;.$$ Then
$$H\setminus\bigcup_{n\in F_1}V_n\subseteq H\cap\bigcup_{n\in F_2}W_n\subseteq G\cap\bigcup_{n\in F_2}W_n\;,$$
so
$$\begin{align*}
H&\subseteq\bigcup_{n\in F_1}V_n\cup\left(G\cap\bigcup_{n\in F_2}W_n\right)\\\\
&\subseteq\bigcup_{n\in F_1}U_n\cup\bigcup_{n\in F_2}U_n\\\\
&\subseteq\bigcup_{n\in F_1\cup F_2}U_n\;,
\end{align*}$$
and 
$$X\setminus H=\bigcup_{n\in F_0}V_n\subseteq\bigcup_{n\in F_0}U_n\;,$$
so
$$X=\bigcup_{n\in F_0\cup F_1\cup F_2}U_n\;.$$
That is, $\{U_n:n\in F_0\cup F_1\cup F_2\}$ is a finite subcover of the $\tau'$-open cover $\mathscr{U}$ of $X$, and $X$ is therefore $\tau'$-compact. Finally, it’s clear that $\tau\subsetneqq\tau'$, since $G\in\tau'\setminus\tau$. This shows that if $\tau$ is maximal countably compact, then every countably compact subset of $X$ is $\tau$-closed.
The converse is easy. Suppose that $\tau$ is not maximal countably compact, and let $\tau'\supsetneqq\tau$ be such that $\langle X,\tau'\rangle$ is countably compact. Fix $V\in\tau'\setminus\tau$, and let $K=X\setminus V$. Then $K$ is $\tau'$-closed and therefore $\tau'$-countably compact. Let $\mathscr{U}\subseteq\tau$ be any countable $\tau$-open cover of $K$; then $\mathscr{U}\subseteq\tau'$, so some finite $\mathscr{U}_0\subseteq\mathscr{U}$ covers $K$, and $K$ is therefore $\tau$-countably compact but not $\tau$-closed.
