[proofplan] The Hausdorff axiom provides disjoint open neighbourhoods $U \ni x$ and $V \ni y$ for any distinct points $x, y$. If a sequence converges to both $x$ and $y$, it is eventually in $U$ and eventually in $V$, placing some term in $U \cap V = \varnothing$ — a contradiction. [/proofplan] [step:Separate the two hypothetical limits using the Hausdorff property] Suppose $(x_n)$ converges to both $x$ and $y$ with $x \neq y$. Since $(X, \tau)$ is Hausdorff, there exist open sets $U$ and $V$ with $x \in U$, $y \in V$, and $U \cap V = \varnothing$. [/step] [step:Derive a contradiction from eventual membership in disjoint open sets] Since $x_n \to x$ and $U$ is an open neighbourhood of $x$, there exists $N_1 \in \mathbb{N}$ such that $x_n \in U$ for all $n \geq N_1$. Since $x_n \to y$ and $V$ is an open neighbourhood of $y$, there exists $N_2 \in \mathbb{N}$ such that $x_n \in V$ for all $n \geq N_2$. For $n \geq \max(N_1, N_2)$, we have $x_n \in U \cap V$. But $U \cap V = \varnothing$, so no such $x_n$ exists — a contradiction. Therefore $x = y$. [guided] The proof has a clean structure: convergence means eventual membership in every open neighbourhood, and the Hausdorff axiom provides two neighbourhoods that cannot be simultaneously inhabited. Together, these force a contradiction. **Setting up the separation:** Suppose $(x_n)$ converges to both $x$ and $y$ with $x \neq y$. The Hausdorff property gives disjoint open sets $U \ni x$ and $V \ni y$ with $U \cap V = \varnothing$. These will serve as the trap. **Convergence to $x$:** Since $x_n \to x$ and $U$ is an open neighbourhood of $x$, there exists $N_1 \in \mathbb{N}$ with $x_n \in U$ for all $n \geq N_1$. **Convergence to $y$:** Since $x_n \to y$ and $V$ is an open neighbourhood of $y$, there exists $N_2 \in \mathbb{N}$ with $x_n \in V$ for all $n \geq N_2$. **The contradiction:** For any $n \geq \max(N_1, N_2)$, the term $x_n$ belongs to both $U$ and $V$, hence to $U \cap V$. But $U \cap V = \varnothing$, so no such $x_n$ exists --- a contradiction. Therefore $x = y$. This argument shows precisely why the Hausdorff axiom is the right separation condition for unique limits: it is the weakest condition guaranteeing that distinct points can be "pulled apart" by disjoint open sets, which is exactly what prevents a sequence from converging to two different points simultaneously. [/guided] [/step]