Here is Prob. 3, Sec. 28, in the book Topology by James R. Munkres, 2nd edition:
Let $X$ be limit point compact.
(a) If $f \colon X \to Y$ 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?
(c) If $X$ is a subspace of the Hausdorff space $Z$, does it follow that $X$ is closed in $Z$?
Here I'll only be attempting Part (a).
My Attempt:
First, we suppose that $f \colon X \to Y$ is a continuous, injective mapping.
Let us put $$ Z \colon= f(X). \tag{Definition A} $$
Then by Theorem 18.2 (e) in Munkres the mapping $g \colon X \to Z$, defined by $$ g(x) = f(x) \ \mbox{ for all } x \in X, \tag{Definition B} $$ is also continuous.
Let $S$ be any infinite subset of $Z = f(X)$. Then the inverse image $f^{-1}(S)$ equals $g^{-1}(S)$ because we have $$ f^{-1}(S) = \{ \ x \in X \ \colon \ f(x) \in S \ \} = \{ \ x \in X \ \colon \ g(x) \in S \ \} = g^{-1}(S). $$ Moreover, this inverse image $f^{-1}(S) = g^{-1}(S)$ is an infinite subset of $X$, by virtue of the fact that $f \colon X \to Y$ is a single-valued mapping, which means that each point of $X$ is mapped under $f$ into one and only one point of $Y$ (and that point in fact is in $Z = f(X)$), and also by virtue of the fact that here $$ S \subset f(X) = \{ \ f(x) \colon x \in X \ \} = \{ \ g(x) \ \colon \ x \in X \ \} = g(X).$$
Now as $X$ is limit point compact and as $f^{-1}(S)$ is an infinite subset of $X$, so $f^{-1}(S)$ has a limit point in $X$; let $p \in X$ be a limit point of $f^{-1}(S)$.
We show that $f(p)$ is a limit point in $f(X)$ of the set $S$.
Let $V$ be any open set in $Z = f(X)$ such that $f(p) \in V$. Then as $g \colon X \to Z$ is continuous, so the inverse image set $g^{-1}(V) = f^{-1}(V)$ is an open set in $X$ such that $p \in g^{-1}(V)$, and since $p$ is a limit point in $X$ of the set $g^{-1}(S) = f^{-1} (S)$, therefore there exists a point $q \in g^{-1}(S) \setminus \{ p \}$ such that $q \in g^{-1}(V)$ also. That is, $$ q \in g^{-1}(V) \cap g^{-1}(S) \qquad \mbox{ and } \qquad q \neq p. $$ Therefore $$ f(q) = g(q) \in V \cap S, $$ and since $f$ (and hence $g$ also) is injective, therefore we also have $f(q) \neq f(p)$ (or $g(q) \neq g(p)$).
Thus we have shown that, for every open set $V$ in $Z = f(X)$ containing $f(p)$, there is a point in $V \cap S$ other than $f(p)$ itself. Thus $f(p)$ is a limit point of set $S$ in $f(X)$.
But $S$ was an arbitrary infinite subset of $f(X)$. Therefore if $f \colon X \to Y$ is a continuous, injective mapping and if $X$ is limit point compact, then so is $f(X)$.
Is this proof correct?
Next, we drop the restriction that $f$ be injective.
If $X$ is compact, then so is $f(X)$, by virtue of Theorem 26.5 in Munkres, and therefore $f(X)$ is also limit point compact, by Theorem 28.1 in Munkres.
Am I right?
Finally we need to consider a topological space $X$ that is limit point compact but not compact.
Let $Y = \{ a, b \}$ with the indiscrete topology, and let $\mathbb{Z} = \{ 0, \pm 1, \pm 2, \pm 3, \ldots \}$ with the discrete topology. Then let us put $$ X \colon = \mathbb{Z} \times Y, $$ and give $X$ the product topology determined by the topologies of $\mathbb{Z}$ and $Y$. This topological space $X$ is limit point compact but not compact. Refer to Example 1, Sec. 28, in Munkres.
Now let us consider the projection map $\pi_1 \colon X \to \mathbb{Z}$ onto the first coordinate, defined by the formula $$ \pi_1( n \times y ) \colon= n \ \mbox{ for all } \ n \times y \in X. $$ This map $\pi_1$ is of course continuous.
However, the image set $\pi_1(X) = \mathbb{Z}$ is not limit point compact. The set $E \colon = \{ 0, \pm 2, \pm 4, \pm 6, \ldots \}$, for example, is an infinite subset of $\mathbb{Z}$ but no $n \in \mathbb{Z}$ can be a limit point of set $E$, because the singleton set $\{ n \}$ is open in the topological space $\mathbb{Z}$, with the discrete topology, and this open set cannot contain any point of the set $E \setminus \{ n \}$.
Is my reasoning here correct and clear enough?
Is my attempt good enough? Or, are there issues?
It seems OK, though, IMHO, overly verbose. The injective part could go, as this has not been asked, but your own addition. But there the domain restricted map $g$ is not needed; I'd write it more like this:
Suppose $A \subseteq f[X]$ is infinite. Then $f^{-1}[A]$ is infinite (or its image $A$ would be finite) so has a limit point $p \in X$ and then $f(p)$ is a limit point of $A$: let $U$ be an open neighbourhood of $f(p)$ (in $Y$); then $f^{-1}[U]$ is an open neighbourhood of $p$ and so contains a point $p' \neq p$ where $p' \in f^{-1}[A]$. It follows that $f(p') \in A \cap U$ and as $f$ is injective, $f(p') \neq f(p)$, and this finishes the proof that $f(p)$ is a limit point of $A$, showing $f[X]$ to be limit point compact.
This is quite a bit shorter than your argument.
The example is fine, I've used it myself in some of my answers here.
If we would have defined $X$ to be $\omega$-limit point compact by demanding that "every infinite subset $A$ of $X$ has an $\omega$-accumulation point (i.e. a point $p \in X$ such that for all open neighbourhoods $O$ of $p$, $O \cap A$ is infinite; in a $T_1$ space a limit point of $A$ is automatically an $\omega$-accumulation point) then $f[X]$ also has that property whenever $X$ has it and $f$ is continuous.
So instead of trying to "fix" $f$ you could have gone for "fixing" $X$ by adding $T_1$-ness. Note that it's not by coincidence that your example space is not $T_1$, not even $T_0$.
The $\omega$-accumulation point property is equivalent, BTW, to the countable compactness of $X$: every countable open cover has a finite subcover.