I'm not sure what you mean by a 'net', but I can show you to get you started.
First, if is Hausdorff then, given any we can find disjoint, open neighbourhoods of and respectively. Then (the last inclusion is due to the disjointness of the neighbourhoods: if then which is wrong by assumption). It follows that is a union of open sets in , so must be open, and therefore is closed.
The converse is similar. If is closed then is open, and so for any we can find an open set such that , and it follows that for some open in . Since , they must be disjoint, and so is Hausdorff.
Hope this helps.