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The open sets are the empty set and subsets whose complement is countable. On an uncountable set, any two nonempty open sets intersect, so the space is not Hausdorff. On a countable underlying set this topology is discrete. The compact subsets of a cocountable space are finite, regardless of the cardinality of the underlying set.

Every subset $A$ of a cofinite topological space is a compact subset. To see the finite reduction directly, take an open cover of a nonempty $A$ by open sets of $X$ and choose one member $U$ containing a point of $A$. Since $U$ has finite complement, the set $A\setminus U$ is finite. Choose a covering member for each of those remaining points. Those members together with $U$ are a finite subcover. The empty set is already compact.

Equivalently, the subspace topology on $A$ is itself cofinite: for finite $F\subseteq X$, the trace of $X\setminus F$ is $A\setminus (F\cap A)$. The argument applies again inside $A$. A concrete open cover $\{X\setminus \{k\}:k\in X\}$ of $X={\mathbb{N}}_{>0}$ needs only the two members $X\setminus \{1\}$ and $X\setminus \{2\}$ when these labels are distinct: each fills the other's single gap.

On $X={\mathbb{N}}_{>0}$, the subset of even integers is compact but not closed, because it is infinite and proper. A compact subset of a Hausdorff space is guaranteed to be closed, whereas compactness without the Hausdorff space assumption does not suffice. Changing the topology changes compactness: the same infinite set with its discrete topology has the singleton open cover, which has no finite subcover. The finite-exception argument is special to finite complements; in the cocountable topology, compact subsets of a cocountable space must instead be finite.

A Hausdorff space is a topological space in which any two distinct points have disjoint open neighborhoods. Let $K$ be a compact subset of a Hausdorff space and $x\notin K$. For each $y\in K$, choose disjoint open neighborhoods $U_y$ of $y$ and $W_y$ of $x$. The $U_y$ cover $K$, so compactness gives a finite subcover $U_{y_1},\dots ,U_{y_m}$. Their paired intersection $W={\bigcap }_{r=1}^m{W}_{y_r}$ is an open neighborhood of $x$ disjoint from $K$. Thus every point of its complement has an open neighborhood in the complement, proving every compact subset of a Hausdorff space is closed. The empty set case is immediate.

In the cocountable topology on an uncountable set, any two nonempty open sets intersect: their combined complements are countable and cannot exhaust $X$. Thus the space is not Hausdorff.

Nevertheless every compact subset is closed here. In fact only finite subsets are compact. To prove this, suppose $K$ is infinite and choose distinct points $d_1,d_2,\dots \in K$. The open setsform an increasing open cover of $X$, and hence of $K$. Any finite selection is contained in $U_N$ for its largest index $N$, which misses $d_N\in K$. So $K$ is not compact. Finite sets are compact in any topological space, and in this cocountable topology they are closed because their complements are open. This proves the claimed behavior of compact subsets of a cocountable space.

A compact space is one for which every open cover has a finite subcover. A Hausdorff space has disjoint open neighbourhoods of any two distinct points.

Let $K$ be a compact subspace of a Hausdorff space $X$, and let $x\notin K$. For each $y\in K$, choose disjoint open neighbourhoods $U_y$ of $y$ and $V_y$ of $x$. Finitely many $U_{y_1},\dots ,U_{y_m}$ cover $K$. Their corresponding intersection $V_{y_1}\cap \cdots \cap V_{y_m}$ is an open neighbourhood of $x$ disjoint from $K$. Thus $X\setminus K$ is open, proving compact subspaces of a Hausdorff space are closed.

In particular, $C_1\cap C_2$ is closed as a subspace of compact $C_1$. A closed subset of a compact space is compact: add its open complement to an open cover, apply compactness to the whole space, and discard that extra open set. Hence $C_1\cap C_2$ is compact.

In the cocountable topology on $\mathbb{R}$, two nonempty open sets always intersect. Their two countable complements have countable union, which cannot exhaust the uncountable real line. Therefore this space is not Hausdorff.

Every finite subset is compact in any topology. Conversely, from any infinite subset $C$ select distinct points $c_1,c_2,\dots$ and defineEach $U_n$ is cocountable and open; the sets increase and their union is all of $\mathbb{R}$. They therefore cover $C$, but a finite selection is contained in some $U_N$, which misses the infinitely many $c_j$ with $j\ge N$. No finite subcover exists. Thus the compact subsets of a cocountable space are