[proofplan] We prove the three-way equivalence by establishing the cycle $(1) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1)$. For $(1) \Rightarrow (3)$, given $f(y_0) \neq g(y_0)$, we separate the images by disjoint open sets in $X$ and pull back to $Y$ via continuity. For $(3) \Rightarrow (2)$, we realise the diagonal as the equaliser of the two canonical projections $\pi_1, \pi_2: X \times X \to X$. For $(2) \Rightarrow (1)$, given $x \neq y$, the point $(x,y)$ lies in the open complement of $\Delta$, and a basic product neighbourhood $U \times V$ of $(x,y)$ missing $\Delta$ forces $U \cap V = \varnothing$. [/proofplan]

[step:Show that the Hausdorff property implies every equaliser is closed — $(1) \Rightarrow (3)$]Assume $(X, \tau)$ is Hausdorff. Let $Y$ be a topological space and let $f, g: Y \to X$ be continuous. Define $E = \{y \in Y : f(y) = g(y)\}$. We show $Y \setminus E$ is open. Let $y_0 \in Y \setminus E$, so $f(y_0) \neq g(y_0)$. Since $X$ is Hausdorff, there exist disjoint open sets $U, V \in \tau$ with $f(y_0) \in U$ and $g(y_0) \in V$. Define $W = f^{-1}(U) \cap g^{-1}(V)$, which is open in $Y$ as a finite intersection of preimages of open sets under continuous maps, and $y_0 \in W$. For any $y \in W$, we have $f(y) \in U$ and $g(y) \in V$. Since $U \cap V = \varnothing$, $f(y) \neq g(y)$, so $y \notin E$. Therefore $W \subset Y \setminus E$, and $Y \setminus E$ is open.[/step]

[guided]Assume $(X, \tau)$ is Hausdorff. Let $Y$ be a topological space and let $f, g: Y \to X$ be continuous. Define the equaliser $E = \{y \in Y : f(y) = g(y)\}$. We show $E$ is closed by showing its complement $Y \setminus E$ is open. The strategy is to show that if $f$ and $g$ disagree at a point, they continue to disagree on an entire neighbourhood. This is precisely the role of the Hausdorff axiom: it provides enough room in $X$ to keep the images apart, and continuity pulls that separation back to $Y$. Let $y_0 \in Y \setminus E$, so $f(y_0) \neq g(y_0)$. Since $X$ is Hausdorff and $f(y_0), g(y_0)$ are distinct points of $X$, there exist disjoint open sets $U, V \in \tau$ with $f(y_0) \in U$ and $g(y_0) \in V$. By continuity of $f$, the preimage $f^{-1}(U)$ is open in $Y$; by continuity of $g$, the preimage $g^{-1}(V)$ is open in $Y$. Define \begin{align*} W = f^{-1}(U) \cap g^{-1}(V). \end{align*} This is open as a finite intersection of open sets, and $y_0 \in W$ since $f(y_0) \in U$ and $g(y_0) \in V$. We verify $W \subset Y \setminus E$. For any $y \in W$, we have $f(y) \in U$ and $g(y) \in V$. Since $U \cap V = \varnothing$, the images $f(y)$ and $g(y)$ are distinct, so $y \notin E$. The disjointness of $U$ and $V$ is the mechanism that ensures $f$ and $g$ remain separated throughout $W$. Since $y_0 \in Y \setminus E$ was arbitrary and we produced an open neighbourhood $W \subset Y \setminus E$ of $y_0$, the set $Y \setminus E$ is open, and $E$ is closed.[/guided]

[step:Realise the diagonal as the equaliser of the two projections — $(3) \Rightarrow (2)$]Assume (3). Consider the product space $Y = X \times X$ with the product topology and the canonical projections \begin{align*} \pi_1: X \times X &\to X, \quad (a, b) \mapsto a, \\ \pi_2: X \times X &\to X, \quad (a, b) \mapsto b. \end{align*} Both $\pi_1$ and $\pi_2$ are continuous by the definition of the product topology. The equaliser of $\pi_1$ and $\pi_2$ is \begin{align*} \{(a, b) \in X \times X : \pi_1(a, b) = \pi_2(a, b)\} = \{(a, b) \in X \times X : a = b\} = \Delta. \end{align*} By assumption (3), this equaliser is closed in $X \times X$.[/step]

[guided]Assume (3): for every topological space $Y$ and every pair of continuous maps $f, g: Y \to X$, the equaliser $\{y \in Y : f(y) = g(y)\}$ is closed. The diagonal $\Delta = \{(x, x) : x \in X\}$ is precisely the locus in $X \times X$ where two natural maps agree — the projections onto each factor. This makes the diagonal an instance of an equaliser. Set $Y = X \times X$ with the product topology, and define $f = \pi_1$ and $g = \pi_2$ where \begin{align*} \pi_1: X \times X &\to X, \quad (a, b) \mapsto a, \\ \pi_2: X \times X &\to X, \quad (a, b) \mapsto b. \end{align*} Both projections are continuous — this is built into the definition of the product topology as the coarsest topology making all projections continuous. The equaliser is \begin{align*} \{(a, b) \in X \times X : \pi_1(a, b) = \pi_2(a, b)\} = \{(a, b) : a = b\} = \Delta. \end{align*} By assumption (3), $\Delta$ is closed in $X \times X$. This establishes (2).[/guided]

[step:Extract disjoint open sets from a product neighbourhood missing the diagonal — $(2) \Rightarrow (1)$]Assume $\Delta$ is closed in $X \times X$. Let $x, y \in X$ with $x \neq y$. Then $(x, y) \notin \Delta$, and since $\Delta$ is closed, the complement $(X \times X) \setminus \Delta$ is open. By the definition of the product topology, there exist open sets $U, V \in \tau$ with \begin{align*} (x, y) \in U \times V \subset (X \times X) \setminus \Delta. \end{align*} In particular, $x \in U$ and $y \in V$. We verify $U \cap V = \varnothing$: if $z \in U \cap V$, then $(z, z) \in U \times V$ and $(z, z) \in \Delta$, contradicting $U \times V \subset (X \times X) \setminus \Delta$. Therefore $U$ and $V$ are disjoint open sets separating $x$ and $y$.[/step]

[guided]Assume the diagonal $\Delta = \{(x, x) : x \in X\}$ is closed in $X \times X$. We show $(X, \tau)$ is Hausdorff. Let $x, y \in X$ with $x \neq y$. The pair $(x, y)$ does not lie on the diagonal (since $x \neq y$), so $(x, y) \in (X \times X) \setminus \Delta$. Since $\Delta$ is closed, this complement is open. The product topology on $X \times X$ has as a basis the collection of sets $U \times V$ where $U, V \in \tau$. Since $(x, y)$ belongs to the open set $(X \times X) \setminus \Delta$, there exist $U, V \in \tau$ with \begin{align*} (x, y) \in U \times V \subset (X \times X) \setminus \Delta. \end{align*} In particular, $x \in U$ and $y \in V$. The inclusion $U \times V \subset (X \times X) \setminus \Delta$ forces $U$ and $V$ to be disjoint: if there existed $z \in U \cap V$, then $(z, z)$ would belong to both $U \times V$ (since $z \in U$ and $z \in V$) and $\Delta$ (since both coordinates equal $z$), contradicting $U \times V \cap \Delta = \varnothing$. Therefore $U \cap V = \varnothing$, and we have produced disjoint open neighbourhoods $U \ni x$ and $V \ni y$. Since $x, y$ were arbitrary distinct points, $(X, \tau)$ is Hausdorff.[/guided]