Prove that topological space $X$ is Hausdorff if and only if limits of all nets in it are unique.
Let $X$ be Hausdorff and $S:D\to X$ be a net in it where $D$ is a directed set. Let $S$ has two limits $x,y\in X$. We need to prove that $x=y$. Suppose $x\neq y$, therefore by Hausdorffness, there exists $U,V$ open sets in $X$ containing $x,y$ respectively such that $U\cap V=\emptyset$. Now by defintion of convergence, there exist $m_1,m_2\in D$ such that for all $n\in D$, $n\geq m_1$ implies $S(n)\in U$ and $n\geq m_2$ implies $S(n)\in V$. Again by definition of directed set, there exists $n\in D$ such that $n\geq m_1$ and $n\geq m_2$. This implies $S(n)\in U\cap V$, a contradiction. Hence $x=y$ and our first part is proved.
Now for the converse, if we take limits of all nets are equal, and WLOG assume $X$ is not Hausdorff. Then for $x,y\in X$, $x\neq y$, we would have two open sets containing $x,y$ respectively which are not disjoint. But I can't get a contradiction this way. How to proceed for this converse part? Any help is appreciated.
Suppose $X$ is not Hausdorff. Then there exist some distinct $x, y \in X$ such that, for all open neighbourhoods $U$ and $V$ of $x$ and $y$ respectively, $U \cap V \neq \emptyset$.
Define the directed set $$A = \{(U, V) : \text{$U, V$ are open and } x \in U, y\in V\},$$ with the partial order $$(U_1, V_1) \le (U_2, V_2) \iff U_1 \supseteq U_2 \text{ and } V_1 \supseteq V_2.$$ Given $(U_1, V_1), (U_2, V_2) \in A$, we can define an upper bound $$(U, V) = \left(U_1 \cap U_2, V_1 \cap V_2\right).$$ Choose (with axiom of choice) a net $(x_{(U, V)})_{(U, V) \in A}$ such that $$x_{(U, V)} \in U \cap V.$$ I claim that $x_{(U, V)} \to x, y$.
For any open neighbourhood $U_0$ of $x$, we have $$(U, V) \ge (U_0, X) \implies x_{(U, V)} \in U \cap V \subseteq U \subseteq U_0,$$ hence $x_{(U, V)} \to x$. A similar argument works for $y$ too.