If $X$ is not limit point compact it has by definition an infinite set $A$ without a limit point. Taking a countably infinite subset $B \subseteq A$ we have that $B$ is closed (as $B' \subseteq A' = \emptyset$) and the same holds for all its subsets, so $B$ has the discrete topology (in its subspace topology; all its subsets are closed in $X$ hence closed in $B$, so all its subsets are open in $B$ as well).
So the question your asking is implied by not being limit point compact, and is in fact equivalent to it. So, if $X$ is a $T_1$ space, this is equivalent to not being "countably compact" (every countable open cover has a finite subcover).
$X = \omega_1$ in the order topology is an example of a non-compact space that is limit point compact, so it has no countable closed discrete subspace.
For metric spaces the equivalence does hold, but not for general spaces.