Showing that a Function is Proper I am trying to solve the following problem:

Let $X$ and $Y$ be two topological spaces. Recall that a function from $X$ to $Y$ is $\it{proper}$ if its pre-image of every compact subset of $Y$ is compact. Prove that if a function from $X$ to $Y$ is closed and its pre-image of each point is compact, then it is proper

I am having trouble doing this. Is this true or is another condition needed, such as $X$ being Hausdorff? I think the following is a counter example that shows that another condition is needed:
Take the interval $[0,1]$ of the real numbers. Take a countable subset $(x_n)_{n \geq 0}$. For each $x_n$ add to this space $n$ points and define a basis for this new space to consist of all open sets of $[0,1]$ with up to a single point replaced by one its copies. Then map this space to $[0,1]$ in the natural way. This natural function seems to be closed, but not proper. And its pre-image of any point is finite; thus, compact. 
Given that this counter example does work, what kind of conidition should be imposed? I will show what line of thought that I have developed to see what extra condition is needed for the proof:
Let $f$ be a closed function from $X$ to $Y$ such that its pre-image on each point is compact. Let $E$ be a compact subset of $Y$. Since the image of $X$ under $f$ is closed, the set $E$ interested with the image of $f$ is compact. This shows that without loss of generality, it can be assumed that $f$ is surjective. This also means that $f$ can be assumed to be open. 
Since $f$ is open, any subset of $X$ that is bijective with $Y$ under $f$ is compact. I then want to try to separate $X$ up or break it down in a step-wise fashion that uses the condition that the pre-image of any point under $f$ is compact. But I cannot get any correspondence between the points that guarantees that such a step-wise method will terminate. 
Does anyone know how to finish this proof with or without another reasonable condition? And if you do not know how to, do you know another way?
 A: Show that $f: X \to Y$ is closed iff (we only need left to right)
$$\forall y \in Y: \forall O \subseteq X: (O \text{ open and } f^{-1}[\{y\}] \subseteq O) \implies \exists U \subseteq Y: U \text{ open }\land  y \in U \land f^{-1}[U] \subseteq O$$
This can be seen by setting $U = Y\setminus f[X\setminus O]$ when $f$ is a closed map. Check the details that it works, or read this answer
Then if $f$ has compact fibres (inverse images of singletons are called fibres) and $C\subseteq Y$ is compact, then take an open cover $U_i, i \in I$ of $f^{-1}[C]$.
For each $y \in C$ finitely many $U_i$ cover its fibre $f^{-1}[\{y\}]$ by compactness of these fibres and  set $O_y$ to be their union, and we get an open neighbourhood $U_y$ of $y$ with $f^{-1}[U_y] \subseteq O_y$  from the above criterion. Now apply compactness of $C$ to the cover of the $U_y$, and then define a finite subcover for the $U_i$. 
I don't understand what your supposed example looks like; you'd need to describe it better, but I think the above proof sketch is watertight. Try it. Introduce some more notation!
