If X is topogy space,Every ultrafilter U on X has single limit, The limit means intersection of clourse of set in U .Does X must has Hausdorff separation?

I guess not,But I don't have an example.

If a topological space $X$ enjoys the property that every ultrafilter $U$ on $X$ has a single limit, must $X$ be a Hausdorff space?

(Ultrafilters here consist of arbitrary subsets (so not necessarily, for example, $z$-sets or closed sets) but the limit of such a $U$ means the intersection of the closures of all its sets.)

I guess not, but I don't have an example.