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Assume that $K $ is compact connected set in a metric space $(M, d)$.

we want to cover $K$ by finite family of open balls $(B_i)_{1\leq i\leq n}$ with equal radius namely, $B_i=B(x_i,r)$; having the property that

$$B_{i-1}\cap B_i \neq \emptyset.$$

Notice that we do not require $B_{j}\neq B_i$ whenever $i \neq j$.

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Start with any finite open cover of balls of radius $r$, call that cover $\mathcal B$.

Pick $B_1 \in \mathcal B$.

Pick $B_2 \in \mathcal B - \{B_1\}$ such that $B_1 \cap B_2 \ne \emptyset$; this is possible since $M$ is connected.

Pick $B' \in \mathcal B - \{B_1,B_2\}$ such that $(B_1 \cup B_2) \cap B' \ne \emptyset$; this is possible since $M$ is connected. If $B' \cap B_2 \ne \emptyset$ then set $B_3 = B'$. On the other hand, if $B' \cap B_1 \ne \emptyset$ then set $B_3 = B_1$ and $B_4 = B'$; notice, you had to back up over $B_1$ again in order to extend the chain.

Continue in this manner inductively, and you'll get all the sets in $\mathcal B$ to be in your chain, perhaps backing up in your chain in order to tack on the next one.

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  • $\begingroup$ From this algorithm initialized by Lee Mosher one can check out that maximum length of the final chain is given by $L_\max = 2^n$. which is computing if at each step one has to repeat the chain from $B_1$ to the last. I think should be a great consequence of this result. $\endgroup$ – Guy Fsone Nov 5 '16 at 16:10

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