Either-Or Topology is Locally Path-Connected

Theorem
Let $T = \left({S, \tau}\right)$ be the either-or space.

Then $T$ is a locally path-connected space.

Proof
Consider the set:
 * $\mathcal B := \left\{{\left\{{x}\right\}: x \in S, x \ne 0}\right\} \cup \left({-1 \,.\,.\, 1}\right)$

Then by Basis for Either-Or Topology, $\mathcal B$ is a basis for $T$.

From Point is Path-Connected to Itself, all $x \in S, x \ne 0$ are path-connected elements of $\mathcal B$.

Finally we consider $\left({-1 \,.\,.\, 1}\right) \in \mathcal B$.

Let $p \in \left({-1 \,.\,.\, 1}\right)$ and consider the mapping $f: \left[{0 \,.\,.\, 1}\right] \to \left({-1 \,.\,.\, 1}\right)$:
 * $\forall x \in \left[{0 \,.\,.\, 1}\right]: f \left({x}\right) = \begin{cases}

0 & : x = 0 \\ p & : x \in \left({0 \,.\,.\, 1}\right] \end{cases}$

Then $f$ is a continuous mapping and so a path from $0$ to $p$.

So $\left({-1 \,.\,.\, 1}\right)$ is a path-connected element of $\mathcal B$.

Hence the result by definition of locally path-connected space.