Finite switching of balls between covering of a compact set

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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.

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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.