[proofplan] The forward direction is a direct application of the Hausdorff separation axiom: if a net converges to two distinct points, any pair of disjoint open neighbourhoods leads to an eventual contradiction with the directedness of the index set. The converse proceeds by contrapositive: if $X$ is not Hausdorff, we construct an explicit net converging to two distinct points by indexing over pairs of neighbourhoods. [/proofplan]

[step:Show that the Hausdorff property forces uniqueness of net limits]Assume $(X, \tau)$ is Hausdorff. Let $(s_\alpha)_{\alpha \in D}$ be a net with $s_\alpha \to x$ and $s_\alpha \to y$, and suppose for contradiction that $x \neq y$. By the Hausdorff property, there exist disjoint open sets $U, V \in \tau$ with $x \in U$, $y \in V$, and $U \cap V = \varnothing$. Since $s_\alpha \to x$, there exists $\alpha_1 \in D$ such that $s_\alpha \in U$ for all $\alpha \succeq \alpha_1$. Since $s_\alpha \to y$, there exists $\alpha_2 \in D$ such that $s_\alpha \in V$ for all $\alpha \succeq \alpha_2$. Since $D$ is directed, there exists $\beta \in D$ with $\beta \succeq \alpha_1$ and $\beta \succeq \alpha_2$. Then $s_\beta \in U \cap V = \varnothing$, a contradiction. Therefore $x = y$.[/step]

[guided]Assume $(X, \tau)$ is Hausdorff, and let $(s_\alpha)_{\alpha \in D}$ be a net converging to both $x$ and $y$. We show $x = y$. Suppose for contradiction that $x \neq y$. The Hausdorff axiom provides disjoint open sets $U$ and $V$ separating $x$ and $y$: $x \in U$, $y \in V$, and $U \cap V = \varnothing$. Convergence $s_\alpha \to x$ means: there exists $\alpha_1 \in D$ such that $s_\alpha \in U$ for all $\alpha \succeq \alpha_1$. Convergence $s_\alpha \to y$ means: there exists $\alpha_2 \in D$ such that $s_\alpha \in V$ for all $\alpha \succeq \alpha_2$. Here is where the directed set axiom is essential. Since $(D, \preceq)$ is directed, we can find $\beta \in D$ with $\beta \succeq \alpha_1$ and $\beta \succeq \alpha_2$. Then $s_\beta \in U$ (from the first tail) and $s_\beta \in V$ (from the second tail), giving $s_\beta \in U \cap V = \varnothing$ — a contradiction. This argument is identical in structure to the [Uniqueness of Limits](/theorems/625) for sequences in Hausdorff spaces. The net generalisation requires no new ideas beyond replacing "there exists $N$" with "there exists $\alpha_0$" — the directedness axiom supplies the common upper bound that $\mathbb{N}$ gives for free via $\max$.[/guided]

[step:Prove the converse by contrapositive — construct a net with two limits when separation fails]We prove the contrapositive: assume $(X, \tau)$ is not Hausdorff and produce a net with two distinct limits. Since $X$ is not Hausdorff, there exist distinct points $x, y \in X$ such that for every pair of open sets $U, V \in \tau$ with $x \in U$ and $y \in V$, the intersection $U \cap V \neq \varnothing$. Define the index set \begin{align*} D := \{(U, V) : U, V \in \tau,\, x \in U,\, y \in V\}, \end{align*} ordered by $(U_1, V_1) \preceq (U_2, V_2)$ if and only if $U_2 \subset U_1$ and $V_2 \subset V_1$ (componentwise reverse inclusion). We verify that $(D, \preceq)$ is directed: given $(U_1, V_1)$ and $(U_2, V_2)$ in $D$, the pair $(U_1 \cap U_2, V_1 \cap V_2)$ belongs to $D$ (since finite intersections of open sets are open, and $x \in U_1 \cap U_2$, $y \in V_1 \cap V_2$) and is an upper bound for both. For each $(U, V) \in D$, choose $s_{(U,V)} \in U \cap V$ (possible since $U \cap V \neq \varnothing$ by the failure of the Hausdorff property). This defines a net $s: D \to X$. The net $(s_{(U,V)})$ converges to $x$: given any open $W$ containing $x$, the pair $(W, X) \in D$, and for any $(U, V) \succeq (W, X)$ we have $U \subset W$, so $s_{(U,V)} \in U \subset W$. The net also converges to $y$: given any open $W$ containing $y$, the pair $(X, W) \in D$, and for any $(U, V) \succeq (X, W)$ we have $V \subset W$, so $s_{(U,V)} \in V \subset W$. Since $x \neq y$, the net has two distinct limits.[/step]

[guided]We prove the contrapositive: if $X$ is not Hausdorff, then some net has two distinct limits. The failure of the Hausdorff property means: there exist $x \neq y$ in $X$ such that no pair of disjoint open neighbourhoods separates them. Formally, for every $U \in \tau$ with $x \in U$ and every $V \in \tau$ with $y \in V$, the intersection $U \cap V \neq \varnothing$. The idea is to build a net that "lives in the intersection of shrinking neighbourhoods of both $x$ and $y$ simultaneously." We index over pairs of neighbourhoods. Define $D := \{(U, V) : U, V \in \tau,\, x \in U,\, y \in V\}$, directed by componentwise reverse inclusion: $(U_1, V_1) \preceq (U_2, V_2)$ iff $U_2 \subset U_1$ and $V_2 \subset V_1$. This is directed because for any two pairs $(U_1, V_1)$ and $(U_2, V_2)$, the pair $(U_1 \cap U_2, V_1 \cap V_2)$ is an upper bound: $U_1 \cap U_2$ is open (finite intersection), contains $x$, and is contained in both $U_1$ and $U_2$; similarly for the second component. For each $(U, V) \in D$, select $s_{(U,V)} \in U \cap V$. This is possible because our non-Hausdorff hypothesis guarantees $U \cap V \neq \varnothing$ for every such pair. Now we verify both convergence claims. For convergence to $x$: let $W$ be an open neighbourhood of $x$. The pair $(W, X)$ belongs to $D$ (since $X$ is open and $y \in X$). For any $(U, V) \succeq (W, X)$, we have $U \subset W$, so $s_{(U,V)} \in U \subset W$. Hence $s_{(U,V)}$ is eventually in $W$. For convergence to $y$: let $W$ be an open neighbourhood of $y$. The pair $(X, W)$ belongs to $D$. For any $(U, V) \succeq (X, W)$, we have $V \subset W$, so $s_{(U,V)} \in V \subset W$. Hence $s_{(U,V)}$ is eventually in $W$. Since $x \neq y$, this net witnesses the failure of unique limits. The construction uses the Axiom of Choice (one selection per pair of neighbourhoods), and the pair-indexing trick generalises the standard sequence-based argument for $T_1$ spaces to arbitrary topological spaces.[/guided]