Not necesssarily locally compact and locally compact difference 
A topological space is said to be locally compact if each point $x\in X$ has at least one neighbourhood which is compact.
  If $f$ is continuous open mapping of a locally compact space $(X,\tau)$ onto a topological space $(Y,\tau_1)$ then $(Y,\tau_1)$ is locally compact.

If $y\in Y$ then there exists a neighbourhood $V$ so that $y\in V$. Suppose there exists at least an $x$ such that $f^{-1}(y)=x$ then $f^{-1}(V)$ contains $U$ that is a compact neighbourhood of $x$. Then $y\in f(U)\subset V$.
In the previous question it was asked:

Prove continuous image of a locally compact space is not necessarily locally compact.

Questions:
1) What is the difference in proof from the present question and the previous one? Why does it change from "not necessarily compact" to "locally compact"?
2) Is my proof right?
Thanks in advance!

As to 1, the difference is that $f$ is now also assumed to be open, when previously we only had continuity. Note that this immediately "kills" the counterexample to that first question, that uses the identity from a discrete space. This  map is only open when the image space also has the discrete topology. 
Your proof is sloppy and unclear, and does not explicitly use the extra assumption of openness, which is a bad sign for a proof.
Better: let $f: X \to Y$ be continuous open and surjective and $X$ locally compact.
Then $Y$ is locally compact: let $y \in Y$ be arbitrary. By ontoness we find $x \in X$ with $f(x)=y$. By local compactness of $X$ there is a compact set $C \subseteq X$ such that $C$ is a neighbourhood of $x$, i.e. there is an open subset $O$ of $X$ with $x \in O \subseteq C$.
Now $$y = f(x) \in f[O] \subseteq f[C]$$
which shows that $f[C]$, which is compact (as $f$ is continuous!) is a neighbourhood of $y$ (as $f[O]$ is open (by openness of $f$!) inside $f[C]$ containing $y$). So $Y$ is locally compact at $y$, and as $y$ was arbitary, everywhere.
Note that we now used all the assumptions on $f$.
