Excluded Point Space is Ultraconnected/Proof 2

Proof
Apart from $S$, every open set of $T$ does not contain $p$, by definition of excluded point space.

So, apart from $\varnothing$, every closed set of $T$ does contain $p$, by definition of closed set.

So every pair of closed sets of $T$ has an intersection which contains at least $p$.

So there are no non-empty disjoint closed sets of $T$.

Hence the result, by definition of ultraconnected.