= Solution
Every <topological subspace> $S\subseteq X$ again has the <cofinite topology>: its nonempty <open subsets> have finite complements in $S$, and every finite exclusion is obtained by intersecting with a cofinite open subset of $X$. As usual, the <empty set> is also open; this exception is implicit in the paper's shorthand definition.
If $S$ is infinite, any two nonempty relatively <open sets> intersect, since their complements are finite. Hence $S$ is <hyperconnected space> and therefore <connected>. If $S$ is finite, its <cofinite topology> is <discrete topology>. It is then <connected> exactly when it has at most one point. Thus
$$
\boxed{S\text{ connected}\iff |S|\leq1\text{ or }S\text{ is infinite}.}
$$
There are also \b[infinite disconnected subsets of the product]. Choose distinct $a,b\in X$ and put $D=\{a,b\}\times X$. This is infinite. The traces on $D$ of the ambient <open sets>
$$
(X\setminus\{b\})\times X,\qquad (X\setminus\{a\})\times X
$$
are respectively $\{a\}\times X$ and $\{b\}\times X$. They are disjoint, nonempty and cover $D$, giving a <separation>. This answers the infinite-subset question, rather than only producing two isolated product points.
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