Let $C$ be an infinite, separable and connected metric space. If $C$ becomes totally disconnected when one of its points $p\in C$ is removed, does every closed ball of $C$ with positive radius and center $p$ always contain an infinite connected subset?

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 "Cantorlike 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. 


I don't know if this answers your question precisely but here's an interesting example. First, let's start with the following: The KnasterKuratowski FanLet $K$ be the KnasterKuratowski fan, also called Cantor's teepee. The space $K$ is defined as follows. Let $C$ be the Cantor set. Let $Q \subset C$ be the set of endpoints of the deleted middlethird intervals. Let $P=C \setminus Q$. We also let $p=(1/2,1/2) \in \mathbb R^2$. Now for each $x \in C$, we let $L_x$ be the line joining $p$ and $x$. Now our space $K$ is the union over all $x \in C$ of the sets
The space $K$ is connected but $K\setminus \{p\}$ is totally disconnected. Also $K$ is punctiform, that is it contains no compact connected T_{2} subspace. A related property of $K$Let $f$ be a continuous function from $K$ to $K$. Let $U$ be some closed connected open set about a point $x \neq p$. Then there exists a closed connected open set $V$ about $p$, which can be written down as the set of all $(x,y)$ with $1/2  \epsilon < y < 1/2$ such that $f(V) \subset U$ since $f$ is continuous. Then $V$ is necessarily homeomorphic to the space $K$. 

