IMO Shortlist

Let $k$ be an arbitrary natural number.

Let $\{m_1,m_2,\ldots{},m_k\}$ be a permutation of $\{1,2,\ldots{},k\}$ such that $a_{m_1} < a_{m_2} < \cdots{} < a_{m_k}$.

Note that we can never have equality since $|a_{m_i} - a_{m_{i+1}}| \ge \frac{1}{m_i+m_{i+1}}$.

Let $\overline{a_ia_j} = |a_i-a_j|$.

By looking at the $a_i$ as a set of intervals on $[0,c]$, it makes sense that $\overline{a_{m_1}a_{m_k}} = \sum \limits_{i=1}^{k-1} \overline{a_{m_i}a_{m_{i+1}}}$.

$\overline{a_{m_i}a_{m_k}} \ge \sum\limits_{i=1}^{k-1} \frac{1}{m_i+m_{i+1}}$.

By the Arithmetic Mean Harmonic Mean Inequality,

$\frac{(a_1+a_2) + (a_2+a_3) + \ldots{} + (m_{k-1}+m_k)}{k-1} \ge \frac{k-1}{\frac{1}{m_1+m_2} + \ldots{} + \frac{1}{m_{k-1}+m_k}}$.

$(m_1+2m_2+\ldots{}+2m_{k-1}+2m_k)\left(\frac{1}{m_1+m_2} + \ldots{} + \frac{1}{m_{k-1}+m_k}\right) \ge (k-1)^2$.

$(\overline{a_{m_1}a_{m_k}})(m_1+2m_2+\ldots{}+2m_{k-1}+m_k) \ge (k-1)^2$.

The right term of the left-hand side is less than $2(m_1+m_2+\ldots{}+m_k)$:

$2\overline{a_{m_1}a_{m_k}}(m_1+m_2+\ldots{}+m_k) > (k-1)^2$

Since $\{m_1,m_2,\ldots{},m_k\}$ is a permutation of $\{1,2,\ldots{},k\}$,

$2\overline{a_{m_1}a_{m_k}} \cdot \frac{k(k+1)}{2} > (k-1)^2$.

$\overline{a_{m_1}a_{m_k}} > \frac{(k-1)^2}{k(k+1)} = \frac{k-1}{k} \cdot \frac{k-1}{k+1} > \left(\frac{k-1}{k+1}\right)^2 = \left(1-\frac{2}{k+1}\right)^2$.

If $\overline{a_{m_1}a_{m_k}} < 1$ for all $k \in \mathbb N$, we can easily find a $k$ such that $\left(1-\frac{2}{k+1}\right)^2 > \overline{a_{m_1}a_{m_k}}$, causing a contradiction.

So $\overline{a_{m_1}a_{m_k}} \ge 1$ for some integers $m_1$, $m_k$.

$|a_{m_1}-a_{m_k}| \ge 1$.

Since both terms are positive, it is clear that at least one of them is greater than or equal to $1$.

So $c \ge 1$, as desired.