If $n = 1$, this problem is trivial; all bounded convex sets are intervals, and no ball centered on the boundary of an interval can disconnect it. Thus, we will consider $n >= 2$.
Since $W$ has a non-empty interior, there must exist a point $p$ and a radius $r$ such that $B_r(p)$ is a subset of the interior of $W$. Since $y_0$ is on the boundary of $W$, we can further say that $B_r(p) \subset (W - \{y_0\})$.
Lemma 1: For any point $q$, there is a sufficiently small radius $r'$ such that removing $B_{r'}(y_0)$ from $W$ leaves $p$ and $q$ connected.
Pick a point $p'$ in the ball $B_r(p)$ which is in general position with $p$, $q$, and $y_0$. In particular, neither $p'p$ nor $p'q$ should pass through $y_0$. This is true for most points in the ball since $n>=2$.
Compute the distance between $y_0$ and each of the lines $pp'$ and $p'q$, and let $r'$ be the lesser of the two. By convexity, $pp'$ and $p'q$ are contained in $W$, and the path given by these two line segments must connect p and q in $W - B_{r'}(y_0)$.
Assume for contradiction that, for any $\delta$, there is some $\delta' < \delta$ s.t. $W - B_{d_i}(y_0)$ is not connected.
Using this, we will define a sequence of points $q_1, q_2, ...$ and values $d_1, d_2, ...$ such that:
- $B_{d_j}(y_0)$ disconnects $q_j$ from $p$
- For $i < j$, $B_{d_j}(y_0)$ does NOT disconnects $q_i$ from $p$
- For $i < j$, the angle between $y_0 q_i$ and $y_0 q_j$ must be at least 60 degrees.
We start by finding a $d_1 < \text{dist}(y_0, p)$ s.t. $W - B_{d_1}(y_0)$ is not connected. Pick $q_1$ to be any point disconnected from $p$ by this.
Lemma 1 tells us that for a sufficiently small radius $r_i$, $B_{d_i}(y_0)$ will NOT disconnect $p$ from $q_i$. Let $s_{j} = \min_{i < j}(r_i)$. Let $t_{j} = \min_{i < j}(\text{dist}(y_0, q_i)) / 2.$ By assumption, we can find $d_{j} < \min(s_j, t_j)$ s.t. $W - B_{d_{j}}(y_0)$ is not connected, so we can pick a point $q_j$ disconnected from $p$ by this. This gives us property (1).
For $i < j$, $d_j \le s_j \le r_i$, so $q_i$ remains connected to $p$. This gives us property (2).
Connectedness is transitive. Since $q_i$ is connected to $p$ but $q_j$ is not, $q_i$ and $q_j$ cannot be connected. Starting here, consider points as vectors, with $y_0$ as the origin. In particular, $|x|$ will denote $\text{dist}(x, y_0)$. We know that $|q_j| \ge d_j$, since by construction $q_j$ was in $W - B_{d_j}(y_0)$. Furthermore, $d_j \le t_j \le |q_i|/2$, so $|q_i| \ge 2d_j$. Now construct vectors $a = d_j/|q_j| * q_j$ and $b = 2d_j/|q_i| * q_i$, noting that $|a| = d_j$ and $|b| = 2d_j$. If Let $\theta$ be the angle between $y_0 q_i$ and $y_0 q_j$. Assume for contradiction that $\theta$ is less than 60 degrees. Then $y_0 a b$ will be an obtuse angle, and the closest point on $a b$ to $y_0$ will be $a$. But because $q_i$ is just $a$ scaled up and $q_j$ is just $b$ scaled up, every point on the line segment $q_i q_j$ is a point on the line segment $a b$ scaled up, so the former line segment can be no closer to $y_0$ than the latter. Thus, all point on $q_i q_j$ have a distance of at least $|a| = d_j$ from $y_0$, which would mean $q_i$ and $q_j$ are connected. That is a contradiction; thus $\theta$ at at least 60 degrees, giving us condition (3).
But there is a cap to the number of vectors in $\mathbb{R}^n$ that can have pairwise angle of at least 60 degrees! Normalizing all the vectors places them on a unit hypersphere, and each vector can be thought of as "claiming" the surface area corresponding an angular change of <30 degrees. This area is a hyperspherical cap. Note that the claimed areas of each vector will be disjoint, since otherwise the two vectors would have angle of less than 30+30 from each other. Both the hypersphere and the hyperspherical cap have a fixed area, so their ratio gives a bound for the number of vectors here.
Thus, the previous construction is impossible - if you let $i$ be any value larger than this cap, then $q_1, ... q_i$ will have pairwise angles of at least 60 degrees, and that can't be the case for any set of $i$ vectors. Thus we have a contradiction.
