A question about the dispersion points of connected metric spaces

Here I construct an example which proves the answer is NO.

Take the KK fan:

Remove its dispersion point at the top. Now you have a Cantor set of lines of rationals/irrationals that cannot be separated horizontally. Stretch this into a "Cantor-like tube" and weave it closer and closer to a point $p$ in the plane while shrinking its diameter and making sure that every loop goes a distance of $7$ away from $p$.

Remove $p$ and you have a hereditarily disconnected space ($\simeq$ KK fan minus its vertex). If $A$ is nonempty and clopen in $X$ then $A$ must snake around the tube forever, so it limits to $p$ and thus $p\in A$. Therefore $X$ is connected. The ball of radius $1$ around $p$ is hereditarily disconnected.

EDIT: I am basically taking the space which consists of the curve below, and the origin $p=(0,0)$ (so $\{0\}\times (0,1]$ is not included). The difference is that instead of weaving a line, I am weaving this "Cantor tube" while shrinking its diameter. In the first case if I remove $p$ then I get something $\simeq [0,\infty)$, whereas in the second case I get something $\simeq$ my Cantor tube.