A locally finite topological space is Alexandrov I'm trying to prove this statement:

A locally finite topological space is Alexandrov.

We know that for every point $x\in X$, we can find a neighbourhood containing only a finite number of points of $X$. 
How can we use that to prove $\cap_{i\in I}A_i $ is open (that is: arbitrary intersections of open sets are open, our definition of an Alexandrov space)?
Thanks for your time.
 A: Let K be a collection of open sets.
If $\cap$K is empty, it is open.
Otherwise, some x in $\cap$K.
Let U be a finite open nhood of x.
Hence $\cap$K = $\cap${ U $\cap$ A : A in K }.
As the elements of K' = { U $\cap$ A : A in K }
are all subsets of U, K' is finite.
Thus $\cap$K = $\cap$K' is open.
A: For every $x$ define $O_x$ to be the unique open set of minimal size that contains $x$. This is well-defined because finite cardinals are well-ordered and if $O_x$ and $O'_x$ would be two distinct open sets of minimal size, then $O_x \cap O'_x$ would have a smaller size (and open and containing $x$) than both, contradicting the minimality. 
Now if $U_i, i \in I$ is any family of open subsets of $X$, and $\bigcap_i U_i$ is non-empty (if it's empty, we're done: the intersection is indeed open) then for $x$ in that intersection, $O_x \subseteq U_i$ for all $i$. For if not, if $O_x \nsubseteq U_i$, then $U_i \cap O_x$ is open and contains $x$ and is strictly smaller in size than $O_x$, again contradicting the minimality of $O_x$. So $O_x \subseteq \bigcap_{i \in I} U_i$ and $x$ is an interior point of the intersection. 
As $x$ was arbitrary, the intersection is open, and $X$ is Aleksandrov.
