**Proof Plan.** The proof for compact [sets](/page/Set) uses the finite intersection property: the complements of the $K_n$ would form an open cover of $K_1$ with no finite subcover, contradicting compactness. The metric-space version uses completeness: choose a point $x_n \in K_n$ for each $n$; the diameter condition forces $\{x_n\}$ to be Cauchy, and completeness gives a [limit](/page/Limit) that lies in every $K_n$. **Step 1 (Compact case — finite intersection property).** Suppose for contradiction that $\bigcap_{n=1}^\infty K_n = \varnothing$. Then $K_1 \subseteq X \setminus \bigcap_{n=1}^\infty K_n = \bigcup_{n=1}^\infty (X \setminus K_n)$. Each $K_n$ is compact in a Hausdorff space, hence closed (by the [Compact Subspaces and Hausdorff Spaces](/theorems/307) theorem), so $X \setminus K_n$ is open. Thus $\{X \setminus K_n\}_{n=1}^\infty$ is an open cover of $K_1$. Since $K_1$ is compact, there exists a finite subcover: $K_1 \subseteq (X \setminus K_{n_1}) \cup \cdots \cup (X \setminus K_{n_m})$ for some $n_1 < \cdots < n_m$. Since the sets are nested ($K_{n_m} \subseteq \cdots \subseteq K_{n_1}$), we have $X \setminus K_{n_1} \subseteq \cdots \subseteq X \setminus K_{n_m}$, so the union equals $X \setminus K_{n_m}$. Therefore $K_1 \subseteq X \setminus K_{n_m}$, i.e. $K_1 \cap K_{n_m} = \varnothing$. But $K_{n_m} \subseteq K_1$ (since $n_m \ge 1$ and the [sequence](/page/Sequence) is decreasing), so $K_{n_m} = \varnothing$, contradicting the hypothesis. **Step 2 (Complete [metric space](/page/Metric%20Space) case).** For each $n \in \mathbb{N}$, choose $x_n \in K_n$. Since $K_n \supseteq K_{n+1} \supseteq \cdots$, we have $x_m \in K_n$ for all $m \ge n$. For $m, m' \ge n$, both $x_m$ and $x_{m'}$ belong to $K_n$, so $d(x_m, x_{m'}) \le \operatorname{diam}(K_n) \to 0$ as $n \to \infty$. Therefore $\{x_n\}$ is a [Cauchy sequence](/page/Cauchy%20Sequence). By completeness, $x_n \to x$ for some $x \in X$. For each fixed $n$, the tail $x_m \in K_n$ for all $m \ge n$, and $K_n$ is closed, so $x = \lim_{m \to \infty} x_m \in K_n$. Therefore $x \in \bigcap_{n=1}^\infty K_n$. **Step 3 (Uniqueness in the metric case).** If $x, y \in \bigcap_{n=1}^\infty K_n$, then $d(x, y) \le \operatorname{diam}(K_n)$ for all $n$, and $\operatorname{diam}(K_n) \to 0$, so $d(x, y) = 0$ and $x = y$. $\blacksquare$