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Define $t_n$ by $t_n = 2^k$ for $2^{k-1} < n \le 2^k$, so that it begins $$1, 2, 4,4, 8,8,8,8, 16,16,16,16,16,16,16,16,32, \dots$$ Then we have $n \le t_n \le 2n$ for all $n$ so your hypothesis is satisfied with $c_1 = 1$, $c_2 = 2$. But there are only about $\log_2 N$ distinct values among $t_1, \dots, t_N$, so $m(\cup^{N}_{n=1}[t_{n}-1,t_{n}+1]) \approx 2 \log_2 N$ and the desired liminf equals zero.