Ok so you obviously have an open set $U$ of the refinement that contains the origin and which is contained in the original origin-centered ball $B$ or radius $1/3$. You're worried that $U$ has to intersect infinitely many refinement sets each of which is confined to its own spine, but thats not necessarily the case. A refinement set, $W$, that intersects $U$ could be contained in $B$, and so $W$ could be the disjoint union of pieces of the interiors of all the spines (subject to the star-finiteness condition). Thus $W$ could "bridge the gap" between $U$ and finitely many of the refinements of the $3/4$ balls. After that you do the same thing again: take a subset, $V$, of $B$ that bridges the gap between $W$ and a few more (finitely many) refinements of the $3/4$ balls, but also such that $V$ is disjoint from $U$ etc.
P.S. On a manifold, you can always take a refinement with an injective pairing, taking an element of the refinement to a superset element of the original cover. This would prohibit the above idea since $W$ would have to find its own superset element of the original cover ($B$ already being taken by $U$). I don't know if there's a name for this property and I don't remember what manifold axioms are used to get it -- it would be interesting and reassuring to see where it fails in the case of the hedgehog space.