Finite switching of balls between covering of a compact set

Let $S \subset \mathbb{R}^n$ be a compact set, and let $\mathfrak{C} = \{B_1, B_2, \cdots, B_m \}$ a finite covering of $S$ with $m$ open balls in $\mathbb{R}^n$. Let $R = \min(r(B_1), \cdots, r(B_m))$ (where $r(B_i)$ denotes the radious of the open ball $B_i$).

Does there exist an absolute constant $c_n$ depending on $n$ and not depending on $m$ such that for any two points $Q_1, Q_2 \subset S$ with $d(Q_1, Q_2) \leq r$, there exists points $P_1, P_2, \cdots, P_{c_n}$ and balls $B_{i_1}, B_{i_2}, \cdots, B_{i_n}$ such that $P_j, P_{j+1} \in B_{i_{j+1}}$ for all $j < n$ and $P_1, Q_1 \in B_{i_1}$ and $P_n, Q_2 \in B_{i_n}$.

PS: If this is true, then that would provide another proof for the following fact: If $f: K \rightarrow \mathbb{R}^n$ with $K$ compact and $f$ continuous, then $f$ is uniformly continuous.

Consider $S = \{0, 1\}$ in $\mathbb{R}$ and the open cover $(-1, 1), (0, 2)$ of $S$. Now for $Q_0 = 0, Q_1 = 1$ in $S$ we have $d(Q_1, Q_2) \leq 1$ and no such sequence of $P_i$'s exist, because in order to be in one of the open balls, one must be either $0$ or $1$ and if two points are in the same ball, they are the same.

• You didn't say that it was connected, only compact. Do you want to edit your question? – 4-ier Aug 23 '18 at 6:54
• Any finite set is compact. – 4-ier Aug 23 '18 at 6:55
• Fix any two points $x$ and $y$. You can take $R$ to be as large as you like and only use balls of radius $R$. Just take the balls, like a linked chain, and walk away from $x$ in a direction opposite that of $y$. Then once you have gotten more than, say, $3R$ away from the both of them, start to make a semi-circle path around to the other side where $y$ is, then go toward $y$. If you want, you can make $S$ to be a path from $x$ to $y$ that lies entirely inside each of these balls. – 4-ier Aug 23 '18 at 8:17
• If $S$ is a connected (non-empty) subset of $\mathbb{R}$, it is then an interval so if $x, y \in S$, $[x,y] \subset S$. For any point $t$ between $x$ and $y$, there is some interval of radius at least $|x-y|$ containing $t$, and hence it contains $x$ or $y$ as well. If one such interval contains both $x$ and $y$ then we are done with only needing one interval. If one $t \in [x,y]$ has an interval containing $x$ and one containing $y$, then we are done with needing two intervals. Else, we get a function $f: [x,y] \to \{x,y\}$ where we map $t$ to the point in one of its intervals. ...Continued.. – 4-ier Aug 23 '18 at 8:32
• Note: because $f(x) = x$ and $f(y) = y$, by the intermediate value theorem, $f$ cannot be continuous. Let $t_0 = \sup \{t \in [x,y]: f(t) = x\}$. If $x < t_0 < y$, then for any $t_0 < t < y, f(t) = y$ and there is a sequence $t_n \to t$ such that $f(t_n) = x$..However, since the interval containing $t_0$ is open, it will contain one such $t_n$ and one such $t > t_0$ and so the intervals of $t_0, t_n,$ and $t$ give you three steps. This gives three intervals needed. (You can check that this is actually attained and is not an over-estimate that cannot happen). .... Continued.... – 4-ier Aug 23 '18 at 8:38