In fact, if $X$ is path-connected and $p \in X$ is a dispersion point, then it is already the case that, either $p$ is contained in no proper closed set (i.e., $\overline{\{p\}} = X$), or $p$ is contained in no proper open set, so the result follows.
Now, to prove the claim, we consider a point $q \notin \overline{\{p\}}$. By path-connectedness, there is a path $f: [0, 1] \to X$ from $q$ to $p$. Then $f^{-1}(\overline{\{p\}})$ is closed in $[0, 1]$ and does not contain $0$, so $r := \min(f^{-1}(\overline{\{p\}})) > 0$. We note that, as $p$ is a dispersion point, any path between two distinct points in $X$ must pass through $p$ (as otherwise the range of the path would be a non-singleton connected subset of $X \setminus \{p\}$, contradicting the latter space’s total disconnectedness). This implies $f(r) = p$ and $f(t) = q$ for all $0 \leq t < r$. By rescaling, this means that $f: [0, 1] \to X$ defined by,
$$f(t) = \begin{cases}
q &, \text{ if }t < 1\\
p &, \text{ if }t = 1
\end{cases}$$
is a path. Now, consider an open set $O$ containing $p$. Then $f^{-1}(O)$ must be open. But $f^{-1}(O)$ is either $\{1\}$, if $q \notin O$; or $[0, 1]$, if $q \in O$. The former case is a contradiction, so any open neighborhood of $p$ must contain $q$. Since $q \notin \overline{\{p\}}$ is arbitrary, this implies any open neighborhood of $p$ must contain $X \setminus \overline{\{p\}}$.
Now, assume $\{p\} \neq \overline{\{p\}}$ and $\overline{\{p\}} \neq X$. I claim that $X \setminus \{p\}$ is then connected, contradicting the assumption that $p$ is a dispersion point. Indeed, let $X \setminus \{p\} = U \cup V$ where $U, V \subset X \setminus \{p\}$ are open in the subspace topology and are disjoint. Pick $q \in \overline{\{p\}} \setminus \{p\}$ and assume WLOG that $q \in U$. We have $U = U’ \setminus \{p\}$ where $U’ \subset X$ is open. Since $q \in \overline{\{p\}}$ and $q \in U’$, we have $p \in U’$. As observed before, this then implies $X \setminus \overline{\{p\}} \subset U’$, so $X \setminus \overline{\{p\}} \subset U$ as well. Now, if $q’$ is another point in $\overline{\{p\}} \setminus \{p\}$, then it cannot be the case that $q’ \in V$, as then the same argument will show $X \setminus \overline{\{p\}} \subset V$ and thus,
$$U \cap V \supset X \setminus \overline{\{p\}} \neq \varnothing$$
contradicting the assumption that $U$ and $V$ are disjoint. Hence, $\overline{\{p\}} \setminus \{p\} \subset U$. Therefore, $U = X \setminus \{p\}$, so $X \setminus \{p\}$ is connected, as claimed.
This implies that there are two possibilities:
- $\{p\} = \overline{\{p\}}$. In this case, as every open neighborhood of $p$ contains $X \setminus \overline{\{p\}} = X \setminus \{p\}$, we have $p$ is contained in no proper open set;
- $\overline{\{p\}} = X$. Of course, this means $p$ is contained in no proper closed set.
Just for completeness, let me add in the proofs that $p$ being contained in no proper closed set or no proper open set implies $X$ is contractible:
If $p$ is contained in no proper closed set, then $F: X \times [0, 1] \to X$ defined by,
$$F(x, t) = \begin{cases}
x &, \text{ if }t = 0\\
p &, \text{ if }t > 0
\end{cases}$$
is a contraction. Indeed, let $K \subset X$ be closed. If $p \notin K$, then $F^{-1}(K) = K \times \{0\}$ is closed. If $p \in K$, then by assumption $K = X$, so $F^{-1}(K) = X \times [0, 1]$ is closed. Thus, $F$ is indeed continuous.
If, instead, $p$ is contained in no proper open set, then $F: X \times [0, 1] \to X$ defined by,
$$F(x, t) = \begin{cases}
x &, \text{ if }t < 1\\
p &, \text{ if }t = 1
\end{cases}$$
is a contraction. Indeed, let $O \subset X$ be open. If $p \notin O$, then $F^{-1}(O) = O \times [0, 1)$ is open. If $p \in O$, then by assumption $O = X$, so $F^{-1}(O) = X \times [0, 1]$ is open. Thus, $F$ is indeed continuous.
