Prob. 3 (a), Sec. 28, in Munkres' TOPOLOGY, 2nd ed: A continuous image of a limit-point compact space is not necessarily limit point compact 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?
 A: 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.

Corollary: if $X$ is $T_1$ and $f:X \to Y$ is continuous, then $f[X]$ is limit point compact. 

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. 
