Equivalence of Definitions of Ultraconnected Space

$(1)$ iff $(2)$
Let $T = \left({S, \tau}\right)$ be ultraconnected by Definition 1:
 * no two non-empty closed sets of $T$ are disjoint.

Let $x, y \in S$.

By Topological Closure is Closed, both $\left\{{x}\right\}^-$ and $\left\{{y}\right\}^-$ are closed.

By hypothesis:
 * $\left\{{x}\right\}^- \cap \left\{{y}\right\}^- \ne \varnothing$

That is, $T = \left({S, \tau}\right)$ is ultraconnected by Definition 2.

Let $T = \left({S, \tau}\right)$ be ultraconnected by Definition 2:
 * $\forall x, y \in S: \left\{{x}\right\}^- \cap \left\{{y}\right\}^- \ne \varnothing$

Let $V_1, V_2$ be closed sets of $T$.

Let $x \in V_1, y \in V_2$.

Then:
 * $\left\{{x}\right\}^- \cap \left\{{y}\right\}^- \ne \varnothing$

But then from Topological Closure of Subset is Subset of Topological Closure we have that:
 * $\left\{{x}\right\}^- \subseteq V_1^-$


 * $\left\{{y}\right\}^- \subseteq V_2^-$

But from Closed Set Equals its Closure $V_1^- = V_1, V_2^- = V_2$.

So from Intersection is Subset:
 * $\left\{{x}\right\}^- \cap \left\{{y}\right\}^- \subseteq V_1$
 * $\left\{{x}\right\}^- \cap \left\{{y}\right\}^- \subseteq V_2$

from which:
 * $V_1 \cap V_2 \ne \varnothing$

As $V_1$ and $V_2$ are arbitrary, it follows that $T$ is ultraconnected by Definition 1.