A topological space is Hausdorff if and only if every net has at most one limit

Suppose $X$ is Hausdorff and a net converges to both $p$ and $q$. If $p\ne q$, take disjoint neighbourhoods $U$ of $p$ and $V$ of $q$; the net is eventually in both, and directedness supplies an index after both thresholds, whose value would lie in $U\cap V$.

Conversely, suppose $X$ is not Hausdorff. Choose distinct $p,q$ for which every neighbourhood of $p$ meets every neighbourhood of $q$, and let $E=\{(U,V,z):U\in\mathcal N(p),\ V\in\mathcal N(q),\ z\in U\cap V\}$, ordered by reverse inclusion in the first two coordinates.

Thus $p=q$, so every net has at most one limit.

The set $E$ is directed: intersect the first two neighbourhood coordinates of two triples and choose a point in their intersection; the resulting triple is above both. The net sending $(U,V,z)$ to $z$ is eventually in every neighbourhood of $p$ and every neighbourhood of $q$, hence converges to both distinct points.

Therefore uniqueness of all net limits forces $X$ to be Hausdorff, and the two implications prove the result.