Past exam of the options II course of the University of Oxford / 2004 / E / E3

A space is Hausdorff if distinct points have disjoint open neighbourhoods. In symbols, for $x\ne y$ there are open $U\ni x$ and $V\ni y$ with $U\cap V=\varnothing$.

A space is compact if every open cover has a finite subcover. The finite-intersection formulation says that every family of closed sets with the finite intersection property has nonempty total intersection.

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

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 the nested intersection is connected.