I'm looking for proof verification/help for the title question. Here is what I have now:
Let $f:X \rightarrow Y$ be continuous with $X$ limit point compact. Let $V$ be an infinite subset of $f(X)$. Then we also have that $f^{-1}(V)$ is also an infinite subset of $X$. Since $X$ is limit point compact there is $x \in X$ such that $x$ is a limit point of $f^{-1}(V)$. Then $x \in \overline{f^{-1}(V)}$ by definition of the closure. Hence since $f$ is continuous, $$ f(x) \in f(\overline{f^{-1}(V)}) \subseteq \overline{f(f^{-1}(V))} \subseteq \overline{V} $$ Then since $f(x) \in \overline{V}$, it is either a limit point of $V$ or an element of $V$. However since is infinite, any neighborhood of $f(x)$ will contain infinitely many points of $V$ in either case. Thus $V$ has a limit point and $f(X)$ is limit point compact.
Your statement does not hold in general.
Here is a counterexample: Take $\mathbb{R}$ with the usual topology and $\{0,1\}$ with the indiscrete topology. Then $X = \mathbb{R} \times \{0,1\}$ with product topology.
$X$ is limit point compact. Take any subset $A\subset X$, then if $(a,0) \in A$, $(a,1)$ is a limit point of $A$ because all neighborhoods of $(a,1)$ also contain $(a,0)$ (by definition of topology of $\{0,1\}$ and product topology). Similarly if $(a,1) \in A$, $(a,0)$ is a limit point of $A$
But the projection map onto $\mathbb{R}$ (just the real line..) is continuous, while the real line is not limit point compact (say $\mathbb{N}$ has no limit point).
For your proof , I don't quite understand "However since is infinite, any neighborhood of $f(x)$ will contain infinitely many points of $V$ in either case". In particular, what if $f(x)$ is an element of $V$?