Ultraconnected Space is Path-Connected

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

Then $T$ is path-connected.

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

Let $a, b \in S$.

Let $p \in \left\{{a}\right\}^- \cap \left\{{b}\right\}^-$ where $\left\{{a}\right\}^-$ is the closure of $\left\{{a}\right\}$.

Such a $p$ can be chosen, as $T$ being ultraconnected guarantees that $\left\{{a}\right\}^- \cap \left\{{b}\right\}^- \ne \varnothing$.

Consider the mapping $f: \left[{0 \,.\,.\, 1}\right] \to X$ such that:
 * $f \left({x}\right) = \begin{cases}

a & : x \in \left[{0 \,.\,.\, \dfrac 1 2}\right) \\ p & : x = \dfrac 1 2 \\ b & : x \in \left({\dfrac 1 2 \,.\,.\, 1}\right] \\ \end{cases}$

Then $f$ is continuous.

The result follows from the definition of path-connectedness.