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!

1) The second question is asking you for an example of topological spaces $X$ and $Y$ and of a function $f\colon X\longrightarrow Y$ such that:


*

*$f$ is continuous;

*$f$ is surjective;

*$X$ is locally compact;

*$Y$ is not locally compact.


2) You did not justify the assertion that $f(U)$ is a neighborhood of $y$. And $f$ doesn't need to have an inverse; therefore, the equality $f^{-1}(y)=x$ makes no sense.
