Denote by $D(Z, r)$ and $S(Z, r)$ the open disc and the circle with center $Z$ and radius $r$, respectively.
Let $A = A_X$ be the set of points that can be reached from $X$ by finite number of jumps (possibly zero). We have to show that $A = \mathbb{R}^2$.

First, we shall prove that $D(X, 2r_X) \subset A$. We may assume that $X = O$ and $r_0 = 1$. Set $a_n = 2^{-n}$, $b_n = 2 - a_n$ and $A_n = \{X \in \mathbb{R}^2 : a_n \le |OX| \le b_n\}$, $n \in \mathbb{N}_0$. It is enough to show that $A_n \subset A$.

We shall proceed by induction. For $n=0$, this follows from the given condition. Suppose that $A_k \subset A$. It suffices to prove that $Y \in A$ for $Y \in [a_{k+1}, a_k) \cup (b_k, b_{k+1}]$ and then apply rotation. Let $f(Z) = f_Y(Z) = |YZ| - r_Z$. It follows by the condition of the problem that $1 - a_{k+1} \le r_{a_k} \le 1 + a_{k+1}$ and $a_{k+1} \le r_{b_k} \le b_{k+1}$. So, if $Y \in [a_{k+1}, a_k)$, then $f(a_k) < a_k - 1 \le 0 < f(-b_k)$, and if $Y \in (b_k, b_{k+1}]$, then $f(b_k) < 0 < 1 - a_{k+1} < f(-a_k)$. Since $A_k$ is a linearly connected set and $f$ is a continuous function, we get that $f(Z) = 0$ for some $Z \in A_k$ and hence $Y \in A$.

Assume now that $A \ne \mathbb{R}^2$. Then $\sup\{r : D(X, r) \subset A\} = R < \infty$. Let $m = \inf\{r_Y : Y \in D(X, R)\}$. Since $r$ is a continuous function, then $m > 0$ and, by the proved above, $D(Y, 2m) \subset A$ for $Y \in D(X, R)$. Hence $D(X, R + 2m) \subset A$, a contradiction.