Let $X$ be limit point compact.
(a) If $f : X → Y$ is continuous, does it follow that $f (X) $is limit point compact?
(b) If $A$ is a closed subset of $X$, does it follow that$ A$ is limit point compact?
for a): yes it true,,as continious image of compact set is compacts
for b) yes it is closed beacuse limit point belong to the set
Is it correct ???
(a) is false (answered several times on this site already):
Let $X$ be the space $\mathbb{N} \times \{0,1\}$ where $\mathbb{N}$ has the (usual) discrete topology and $\{0,1\}$ has the indiscrete topology, and $X$ has the product topology of these. Let $f$ be the projection on the first coordinate. ($X$ is defined in exercise 1 in that section, setting it up for you.)
$X$ is limit point compact, because if $A \subseteq X$ and $(n,i) \in A$, then $(n,i')$ is a limit point of $A$ (where $i' = 1$ if $i=0$ and vice versa), because any neighbourhood of $(n,i')$ contains $\{n\} \times \{0,1\}$ and thus $(n,i) \in A \setminus \{(n,i')\}$.
But $f[X]$ is an infinite discrete space, so not limit point compact, as no subset of a discrete space has a limit point.
(b) is true: if $X$ is limit point compact, and $A$ is closed, then any infinite $B \subseteq A$ has a limit point in $X$ and this limit point is in $A$, as $B' \subseteq A' \subseteq A$.
(c) is false: $S_\Omega$ (the first uncountable ordinal in the order topology) is limit point compact but not closed in $S_\Omega \cup \{S_\Omega\}$, the successor ordinal (which Munkres denotes $\overline{S_\Omega}$, I believe).
(d)((a) - (c) under the extra Hausdorff condition)) (c) remains false as the same example works (both are Hausdorff spaces). (b) remains true, and (a) becomes true, because ($T_1$ is already enough) then a limit point is the same as an $\omega$-limit point (every neighbourhood contains infinitely many points of the set). Those proofs are not hard and can be found on this site as well.