Particular Point Space is Locally Path-Connected

Theorem
Let $T = \left({S, \tau_p}\right)$ be a particular point space.

Then $T$ is locally path-connected.

Proof
Consider the set $\mathcal B$ defined as:
 * $\mathcal B = \left\{{\left\{{x, p}\right\}: x \in S}\right\}$

Then $\mathcal B$ is a basis for $T$.

Now consider the open set $\left\{{p, q}\right\} \in \mathcal B$.

Let $\mathbb I$ be the closed unit interval in $\R$.

Let $f: \mathbb I \to S$ be the mapping defined as:
 * $\forall x \in \mathbb I: f \left({x}\right) = \begin{cases}

p & : x \in \left[{0 \,.\,.\, 1}\right) \\ q & : x = 1 \end{cases}$

Suppose $U \in \tau_p$.

Then either $q \in U$ or $q \notin U$.

If $q \in U$ then $f^{-1} \left({U}\right) = \left[{0 \,.\,.\, 1}\right]$ which is open in $\mathbb I$ because $\left[{0 \,.\,.\, 1}\right] = \mathbb I$.

If $q \notin U$ then $f^{-1} \left({U}\right) = \left[{0 \,.\,.\, 1}\right)$ which is half open in $\R$ but open in $\mathbb I$.

So $f: \mathbb I \to S$ is a continuous mapping and so a path from $p$ to $q$.

As $\left\{{p, q}\right\}$ is any element of $\mathcal B$ it follows that $T$ is locally path-connected.