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The open sets $U$ and $X\setminus C_{\lambda }$ cover $X$. A finite subcover exists by compactness, and taking complements yieldsfor a finite index set. This is the finite intersection characterization of compactness in a neighborhood form; the finite index set may be empty when $U=X$.

The Šura-Bura lemma states that in a compact Hausdorff space the connected component $C(x)$ of a point equals its quasi-component, the intersection of all clopen neighborhoods of that point. It also follows that each component has a neighborhood basis of clopen sets. Here is a proof of both assertions.

Let $Q$ be the intersection of the clopen neighborhoods of $x$. Every connected set containing $x$ lies in each such neighborhood, so $C(x)\subseteq Q$. The set $Q$ is closed and compact. If $Q$ were disconnected, write it as the disjoint union of nonempty relatively closed sets $A,B$, choosing $x\in A$. Both are compact and closed in $X$. Normality gives disjoint open sets $O_A,O_B$ containing them. By the finite-intersection result, a finite intersection $V$ of clopen neighborhoods of $x$ satisfies $Q\subseteq V\subseteq O_A\cup O_B$. Then $V\cap O_A$ is clopen in $V$, hence clopen in $X$, and contains $x$ while excluding $B$. This contradicts the definition of $Q$. Thus $Q$ is connected and equals $C(x)$. Given any open neighborhood of the component, the same finite-intersection argument supplies a clopen neighborhood inside it. This proves the neighborhood-basis form as well.

For an explicit noncompact counterexample to the Šura-Bura lemma, put $p=(0,0)$, $q=(1,0)$, and ${\epsilon }_m=1/(m+2)$ for $m\ge 1$. Let $A_m$ be the polygonal arc with successive verticesand give $Z=\{p,q\}\cup {\bigcup }_{m\ge 1}{A}_m$ its subspace topology from the plane. The arcs are pairwise disjoint: a higher arc has its vertical sides outside the horizontal span of any lower arc. Each $A_m$ is compact and relatively open in $Z$, so is clopen. For relative openness, all sufficiently high arcs have vertical sides close to $x=0,1$ and horizontal sides above a fixed bounded neighborhood of $A_m$; the remaining finitely many compact disjoint arcs have positive distance from it.

A connected subset containing $p$ cannot meet any of these clopen arcs, so it lies in $\{p,q\}$ and is just $\{p\}$. But any clopen neighborhood of $p$ contains the left endpoints of all sufficiently high arcs. Connectedness of each arc forces the entire arc into that neighborhood, and closedness then forces the limit $q$ of their right endpoints into it. Conversely the clopen set $Z\setminus A_m$ excludes every point of $A_m$. ThereforeThe compact set containing the endpoints has lost the connection supplied by arcs escaping to arbitrarily large height; compactness of the ambient space is essential to the lemma.

Now let $D={\bigcap }_n{C}_n$ for the decreasing compact connected sets in the compact Hausdorff setting. Every finite intersection is one of the nonempty $C_n$, so the finite intersection property gives $D\ne \varnothing$. It is closed and hence compact. If it were disconnected, its two nonempty compact pieces would have disjoint open neighborhoods $U,V$ in $X$. Since $D\subseteq U\cup V$, the initial finite-intersection result and the decreasing property give $C_N\subseteq U\cup V$ for some $N$. This set meets both neighborhoods because it contains $D$, contradicting its connectedness. ThusThe argument does not need connectedness of the whole ambient space.

For the requested failure without ambient compactness, add the central ray $L=\{(1/2,t):t\ge 1\}$ to the previous example and put $X=Z\cup L$. For $n\ge 1$ defineEach graph formed by the displayed arcs and ray is connected: each arc meets the ray at its top. Adding $p,q$, which are limit points of the graph, preserves connectedness because the closure of a connected set is connected. Its relative closure in $X$ adds exactly these two points, so $C_n$ is closed in $X$. To check that no omitted point is a further limit, a bounded neighborhood away from the lines $x=0,1$ meets only finitely many arcs; the deleted initial arcs are disjoint compact sets, and the deleted ray segment lies strictly below the remaining ray. All other planar limit points on $x=0,1$ are outside $X$.

The sets are decreasing and nonempty, and $X=C_1$ is connected and Hausdorff. Every individual arc and every point of finite height on the ray disappears from sufficiently late sets. Consequentlya nonempty compact disconnected set. Reindexing by $n+1$ gives the source's indexing by $\omega$. The graph is unbounded, so the ambient subspace is not compact. This is an explicit decreasing closed connected sets with disconnected compact intersection example.