Continuous image of a locally compact space is not necessarily locally compact 
A topological space is said to be locally compact if each point $x\in X$ has at least one neighbourhood which is compact.
  Prove continuous image of a locally compact space is not necessarily locally compact.

Attempted solution:
I was able to find a counter example I believe:
$g:(X',\tau')\to(Y',\tau'')$ where $\tau$ is the discrete topology and $\tau_2$ is the indiscrete topology where $Y'$ is infinite, hence it is not compact(there is no finite open set but $\emptyset$).
However I am striving to see why the image is not necessarily locally compact.
Admitting $f:(X,\tau)\to (Y,\tau_1)$ is a continuous function. Then if $y\in Y$ then $f^{-1}(y)=x$ since $(X,\tau)$ is locally compact then  $x\in U$ such that $U$ is compact. Them image of a compact set is a compact set so $y\in f(U)$ so that $f(U)$ is compact. So the image is locally compact.
Question:
1) Since the author states not "necessarily compact". What am I missing?
2) Is my counterexample right?
Thanks in advance!
 A: Your counterexample is wrong because the indiscrete topology is locally compact in your definition, but it can be fixed:
If $\tau_d$ is the discrete topology on a set $X$ then $f(x)=x$ is continuous 
as a map between $(X,\tau_d)$ to $(X,\tau)$ where $\tau$ is any topology on $X$ we like. This observation does hold.
Also, $(X,\tau_d)$ is locally compact as every point $x$ has the compact neighbourhood $\{x\}$ in $\tau_d$.
So let $(X,\tau)$ be any non-locally compact topology. (Like $\mathbb{Q}$ in the standard topology inherited from $\mathbb{R}$ (or its order)) and use that special case. The same "example template" can also be used for metrisable spaces, locally connected spaces etc.
So you then have shown that the continuous image of a locally compact space is not necessarily locally compact (you have at least one concrete example where this is not the case). E.g. the continuous image of a compact space is "necessarily compact", because there is a theorem that says so. 
A: Your example is wrong because every subset is compact in indiscrete topology. Also in the proof that follows $f(U)$ is compact but it need not be a neighborhood of $y$.
