[proofplan] We first establish that $K := \bigcap_{n=1}^\infty K_n$ is non-empty and compact using the finite intersection property. For connectedness, we argue by contradiction: if $K$ were disconnected, a separation $K = A \cup B$ by disjoint closed sets could be extended to disjoint open sets $U \supset A$ and $V \supset B$ (using the Hausdorff property and compactness). The sets $K_n \setminus (U \cup V)$ would form a decreasing sequence of compact sets; if all non-empty, their intersection — a subset of $K$ disjoint from $U \cup V$ — would be non-empty, contradicting $K \subset U \cup V$. So some $K_n \subset U \cup V$, but then $K_n$ would be the union of the disjoint non-empty open sets $K_n \cap U$ and $K_n \cap V$, contradicting the connectedness of $K_n$. [/proofplan]

[step:Show that $K$ is non-empty and compact]Set $K := \bigcap_{n=1}^\infty K_n$. The family $\{K_n\}_{n=1}^\infty$ has the finite intersection property: for any finite subcollection $K_{n_1}, \ldots, K_{n_m}$, the intersection $K_{n_1} \cap \cdots \cap K_{n_m} = K_{\max(n_1, \ldots, n_m)}$ is non-empty (since every $K_n$ is non-empty and the sequence is decreasing). Since $K_1$ is compact and each $K_n$ is a closed subset of $K_1$ (compact subsets of a Hausdorff space are closed, by [Compact Subspaces and Hausdorff Spaces](/theorems/307)), the finite intersection property guarantees $K = \bigcap_{n=1}^\infty K_n \neq \varnothing$. The set $K$ is compact because it is a closed subset of the compact space $K_1$: each $K_n$ is closed in $X$ (again by [Compact Subspaces and Hausdorff Spaces](/theorems/307), since $X$ is Hausdorff and $K_n$ is compact), so $K = \bigcap_{n=1}^\infty K_n$ is an intersection of closed sets, hence closed. A closed subset of the compact space $K_1$ is compact (by part (1) of [Compact Subspaces and Hausdorff Spaces](/theorems/307)).[/step]

[guided]Before proving connectedness, we need to confirm that $K$ is non-empty and compact. Both facts rely on a single key observation: in a Hausdorff space, compact sets are closed. For non-emptiness, we use the characterisation of compactness via the finite intersection property: a topological space is compact if and only if every family of closed subsets with the finite intersection property has non-empty total intersection. The sets $K_n$ are closed in $K_1$ (since they are compact subsets of the Hausdorff space $X$, hence closed in $X$, hence closed in the subspace $K_1$). The nesting $K_1 \supset K_2 \supset \cdots$ ensures that any finite intersection $K_{n_1} \cap \cdots \cap K_{n_m} = K_{\max(n_1, \ldots, n_m)} \neq \varnothing$. By compactness of $K_1$, the total intersection $K = \bigcap_{n=1}^\infty K_n$ is non-empty. For compactness of $K$: each $K_n$ is compact in $X$, hence closed in $X$ (since $X$ is Hausdorff). The intersection $K = \bigcap K_n$ is therefore closed in $X$, hence closed in $K_1$. By part (1) of [Compact Subspaces and Hausdorff Spaces](/theorems/307), a closed subset of a compact space is compact, so $K$ is compact.[/guided]

[step:Assume for contradiction that $K$ is disconnected and separate it by open sets]Suppose $K$ is not connected. Then by the [Equivalent Characterisations of Connectedness](/theorems/294), there exist non-empty disjoint open sets $A_0, B_0$ in the subspace topology of $K$ with $K = A_0 \cup B_0$. Since $K$ is compact and Hausdorff (as a subspace of a Hausdorff space), $A_0$ and $B_0$ are also closed in $K$ (each is the complement of the other), hence compact (closed subsets of a compact space are compact). We now separate $A_0$ and $B_0$ by open sets in $X$. Since $X$ is Hausdorff and $A_0, B_0$ are disjoint compact subsets, we can separate them: for each $a \in A_0$ and $b \in B_0$, the Hausdorff property provides disjoint open sets $U_{a,b} \ni a$ and $V_{a,b} \ni b$. For fixed $a$, the collection $\{V_{a,b}\}_{b \in B_0}$ covers $B_0$. By compactness of $B_0$, finitely many suffice: $B_0 \subset V_{a,b_1} \cup \cdots \cup V_{a,b_m}$. Set $U_a := U_{a,b_1} \cap \cdots \cap U_{a,b_m}$ and $V_a := V_{a,b_1} \cup \cdots \cup V_{a,b_m}$. Then $U_a$ is open with $a \in U_a$, $V_a$ is open with $B_0 \subset V_a$, and $U_a \cap V_a = \varnothing$. Now $\{U_a\}_{a \in A_0}$ covers $A_0$. By compactness of $A_0$, finitely many suffice: $A_0 \subset U_{a_1} \cup \cdots \cup U_{a_k}$. Set \begin{align*} U := U_{a_1} \cup \cdots \cup U_{a_k}, \qquad V := V_{a_1} \cap \cdots \cap V_{a_k}. \end{align*} Then $U$ and $V$ are open in $X$, $A_0 \subset U$, $B_0 \subset V$, and $U \cap V = \varnothing$.[/step]

[guided]This step is the technical heart of the argument. We need to pass from a separation of $K$ in the subspace topology to a separation by open sets of the ambient space $X$. Suppose $K = A_0 \cup B_0$ with $A_0, B_0$ non-empty, disjoint, and open in the subspace topology of $K$. Since $K = A_0 \cup B_0$ is a disjoint union and both sets are open in $K$, each is the complement of the other in $K$, hence also closed in $K$. Since $K$ is compact, $A_0$ and $B_0$ — being closed subsets of a compact space — are themselves compact. Now we must find disjoint open sets $U \supset A_0$ and $V \supset B_0$ in $X$. This is a standard compactness argument in Hausdorff spaces, proceeding in two stages: **Stage 1: Separate a single point from $B_0$.** Fix $a \in A_0$. For each $b \in B_0$, since $a \neq b$ and $X$ is Hausdorff, there exist disjoint open sets $U_{a,b} \ni a$ and $V_{a,b} \ni b$. The collection $\{V_{a,b}\}_{b \in B_0}$ is an open cover of $B_0$. By compactness of $B_0$, there exist $b_1, \ldots, b_m$ with $B_0 \subset V_{a,b_1} \cup \cdots \cup V_{a,b_m}$. Setting $U_a = \bigcap_{j=1}^m U_{a,b_j}$ and $V_a = \bigcup_{j=1}^m V_{a,b_j}$, we have $a \in U_a$ (a finite intersection of open sets), $B_0 \subset V_a$ (a union of open sets), and $U_a \cap V_a = \varnothing$ (since $U_a \subset U_{a,b_j}$ and $U_{a,b_j} \cap V_{a,b_j} = \varnothing$ for each $j$). **Stage 2: Separate $A_0$ from $B_0$.** The collection $\{U_a\}_{a \in A_0}$ covers $A_0$. By compactness, there exist $a_1, \ldots, a_k$ with $A_0 \subset U_{a_1} \cup \cdots \cup U_{a_k}$. Setting $U = \bigcup_{i=1}^k U_{a_i}$ and $V = \bigcap_{i=1}^k V_{a_i}$, we get disjoint open sets with $A_0 \subset U$ and $B_0 \subset V$. Why the asymmetry between union and intersection? For $U$, we take a union to ensure $A_0 \subset U$, and for $V$, we take an intersection to ensure disjointness: $U_{a_i} \cap V_{a_i} = \varnothing$ for each $i$, so $U_{a_i} \cap V = \varnothing$ (since $V \subset V_{a_i}$), and therefore $U \cap V = \varnothing$.[/guided]

[step:Derive a contradiction from the connectedness of each $K_n$]We have disjoint open sets $U, V$ in $X$ with $A_0 \subset U$, $B_0 \subset V$, and $K = A_0 \cup B_0 \subset U \cup V$. Consider the sets $L_n := K_n \setminus (U \cup V)$ for each $n \in \mathbb{N}$. Each $L_n$ is closed in $X$: $K_n$ is compact (hence closed in the Hausdorff space $X$), and $U \cup V$ is open, so $L_n = K_n \cap (X \setminus (U \cup V))$ is the intersection of two closed sets. The sequence $(L_n)$ is decreasing since $(K_n)$ is decreasing. [claim:Some $L_N$ is empty] There exists $N \in \mathbb{N}$ such that $L_N = K_N \setminus (U \cup V) = \varnothing$, i.e. $K_N \subset U \cup V$. [/claim] [proof] Suppose for contradiction that every $L_n$ is non-empty. Each $L_n$ is a closed subset of the compact set $K_1$, hence compact. The family $\{L_n\}_{n=1}^\infty$ is a decreasing sequence of non-empty compact sets, so by the finite intersection property (applied as in the first step), $\bigcap_{n=1}^\infty L_n \neq \varnothing$. But \begin{align*} \bigcap_{n=1}^\infty L_n = \bigcap_{n=1}^\infty \big(K_n \setminus (U \cup V)\big) = \left(\bigcap_{n=1}^\infty K_n\right) \setminus (U \cup V) = K \setminus (U \cup V) = \varnothing, \end{align*} since $K \subset U \cup V$. This is a contradiction. [/proof] Fix $N$ such that $K_N \subset U \cup V$. Then \begin{align*} K_N = (K_N \cap U) \cup (K_N \cap V). \end{align*} The sets $K_N \cap U$ and $K_N \cap V$ are open in the subspace topology of $K_N$ and disjoint (since $U \cap V = \varnothing$). Neither is empty: $A_0 \subset K \subset K_N$ and $A_0 \subset U$ give $A_0 \subset K_N \cap U$, so $K_N \cap U \supset A_0 \neq \varnothing$; similarly $K_N \cap V \supset B_0 \neq \varnothing$. Therefore $K_N \cap U$ and $K_N \cap V$ form a separation of $K_N$ into two non-empty disjoint open subsets, contradicting the connectedness of $K_N$. This contradiction shows that $K$ is connected.[/step]

[guided]We have found disjoint open sets $U \supset A_0$ and $V \supset B_0$ in $X$ with $K \subset U \cup V$. Now we connect this back to the sets $K_n$. The key question: does some $K_n$ fit inside $U \cup V$? If so, then $K_n$ would be covered by two disjoint open sets, each meeting $K_n$ non-trivially (since $A_0 \subset K \subset K_n$ and $A_0 \subset U$, and similarly for $B_0$ and $V$). This would disconnect $K_n$, contradicting the hypothesis. To show some $K_n$ fits inside $U \cup V$, consider the "leftovers" $L_n := K_n \setminus (U \cup V)$. These are the points of $K_n$ not covered by $U \cup V$. We need to show $L_N = \varnothing$ for some $N$. If every $L_n$ were non-empty, the sequence $(L_n)$ would be a decreasing sequence of non-empty compact sets (each $L_n$ is closed in $K_1$, hence compact). By the finite intersection property, $\bigcap L_n \neq \varnothing$. But $\bigcap L_n = K \setminus (U \cup V) = \varnothing$, since $K = A_0 \cup B_0 \subset U \cup V$. Contradiction. So there exists $N$ with $K_N \subset U \cup V$. Then $K_N = (K_N \cap U) \cup (K_N \cap V)$ is a partition of $K_N$ into two disjoint sets, each open in the subspace topology of $K_N$. Both are non-empty: $K_N \cap U \supset A_0 \neq \varnothing$ (since $A_0 \subset K \subset K_N$ and $A_0 \subset U$), and $K_N \cap V \supset B_0 \neq \varnothing$ by the same reasoning. This gives a disconnection of $K_N$, contradicting the assumption that each $K_n$ is connected. The contradiction shows that the original assumption — that $K$ is disconnected — must be false. Therefore $K$ is connected. It is worth noting that the Hausdorff hypothesis is used in two essential ways: (1) to guarantee that compact sets are closed (so that the finite intersection property yields non-empty intersections), and (2) to separate the compact sets $A_0$ and $B_0$ by disjoint open sets $U$ and $V$.[/guided]