When $C^* (X) = C(X)$, $X$ is said to be pseudocompact. Every compact space is pseudocompact.
my questions:
1: IS $X$ pseudo-compact if only if $f[X] $ is compact for $f$ in $C^{*}(X)$?
$f[X] = \{ fx : x \in X \} $
We know that a Hausdorff space is countably compact if only if every infinite set has a limit point.
2:Let $X$ be a Hausdorff space. If, of any two disjoint closed sets, at least one is compact, or even countably compact, is $X$ countably compact?
Fact: If $X$ is pseudocompact, and $f: X \to Y$ is continuous then $f[X]$ is pseudocompact. The proof is immediate.
Fact 2: for metric spaces, pseudocompactness is equivalent to compactness.
These facts solve your first question.
As to the second: if $X$ is not countably compact (and $T_2$), $X$ has a countable closed and discrete set $A$. Split $A$ into two such sets. Then neither is (countably ) compact. This settles 2.