= Solution
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 $U_{y_1},\ldots,U_{y_m}$ 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 \b[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 \b[$C_1\cap C_2$ 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 \b[this space is not Hausdorff].
Every finite subset is compact in any topology. Conversely, from any infinite subset $C$ select distinct points $c_1,c_2,\ldots$ and define
$$
U_n=\mathbb R\setminus\{c_j:j\geq n\}.
$$
Each $U_n$ 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 $j\geq N$. No finite subcover exists. Thus the <compact subsets of a cocountable space> are
$$
\boxed{\text{exactly the finite subsets}.}
$$
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