Non-Trivial Ultraconnected Space is not T1

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space which is ultraconnected.

If $S$ has more than one element, then $T$ is not a $T_1$ (Fréchet) space.

That is, if $T$ is a $T_1$ (Fréchet) space with more than one element, it is not ultraconnected.

Proof
$T = \left({S, \tau}\right)$ be ultraconnected.

Thus by definition:
 * $(1): \quad \forall x, y \in S: \left\{{x}\right\}^- \cap \left\{{y}\right\}^- \ne \varnothing$

Let $a, b \in S$ such that $a \ne b$.

$T$ is a $T_1$ (Fréchet) space.

By definition of $T_1$ Space, $\left\{{a}\right\}$ and $\left\{{b}\right\}$ are closed.

From Closed Set Equals its Closure we have that $\left\{{a}\right\}^- = \left\{{a}\right\}$ and $\left\{{b}\right\}^- = \left\{{b}\right\}$.

It immediately follows that:
 * $\left\{{a}\right\}^- \cap \left\{{b}\right\}^- = \varnothing$

But that contradicts $(1)$ above.

The result follows by Proof by Contradiction.