= Solution
The nested closed sets have the <finite intersection property>: the intersection of finitely many is the one with the largest index and is nonempty. Compactness gives $\boxed{Y_\infty\ne\varnothing}$.
For an open neighbourhood $W$ of the intersection, suppose no $Y_n$ were contained in $W$. Then $Z_n=Y_n\cap(X\setminus W)$ would be nested, nonempty and closed. The same argument gives a point in all $Z_n$, contradicting $Y_\infty\subseteq W$. Hence
$$
\boxed{Y_n\subseteq W\text{ for some }n.}
$$
For <connectedness>, suppose $Y_\infty=A\cup B$ were a separation into two nonempty relatively closed sets. Since $Y_\infty$ is <compact>, $A,B$ are disjoint <compact> subsets of $X$. The supplied separation fact gives disjoint open neighbourhoods $U_A,U_B$. Their union contains $Y_\infty$, so some $Y_n$ lies in that union. But $Y_n$ meets both neighbourhoods, since it contains $A$ and $B$. Its intersections with the two neighbourhoods are then a separation, contradicting its <connectedness>. Thus \b[the nested intersection is <connected>].
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