Equivalence of Definitions of Ultraconnected Space/1 iff 3

Closed Sets Intersect implies Closed Sets are Connected
Let $T = \left({S, \tau}\right)$ be ultraconnected in the sense that:
 * no two non-empty closed sets of $T$ are disjoint.

Let $F \subseteq S$ be an arbitrary closed set of $T$.

$F$ is not connected.

Then there exist non-empty closed set $G, H$ in $F$ that are disjoint (and whose union is $F$).

By Closed Set in Closed Subspace, $G$ and $H$ are closed in $T$.

Because $G \cap H = \varnothing$, $T$ is not ultraconnected.

This is a contradiction.

Thus $F$ is connected.

As $F$ is arbitrary, this applies to all closed set of $T$.

Thus $T = \left({S, \tau}\right)$ is ultraconnected in the sense that:
 * every closed set of $T$ is connected..

Closed Sets are Connected implies Closed Sets Intersect
Let $T = \left({S, \tau}\right)$ be ultraconnected in the sense that:
 * every closed set of $T$ is connected.

Let $G$ and $H$ be closed sets of $T$.

Then their union $G \cup H$ is closed in $T$.

By assumption, $G \cup H$ is connected.

By Closed Set in Closed Subspace, $G$ and $H$ are closed sets of $G \cup H$.

Because $G \cup H$ is connected, $G \cap H$ is non-empty.

Because $G$ and $H$ were arbitrary, this applies to intersection of all such closed sets

Thus $T = \left({S, \tau}\right)$ is ultraconnected in the sense that:
 * no two non-empty closed sets of $T$ are disjoint.

Also see

 * Closed Subset of Ultraconnected Space is Ultraconnected
 * Space is Irreducible iff Open Subsets are Connected