Let {${\alpha_n}$} be a sequence in $\mathbb R$ satisfying $|\alpha_n - \alpha_{n+1}|\geq c$ for some $c > 0$ and all $n \in \mathbb N$. Prove that the sequence {$\alpha_n$} diverges.
I've looked up a few examples on how to show a sequence diverges, where they talk about showing that one variation of the sequence goes to one limit, and another variation goes to a different limit, but I wasn't sure how to do that with this.
Well you don't need to use contradiction. But you need to use negation or maybe a definition given by a professor for a sequence to diverge but I've found that teachers tend to omit this. A sequence $(a_n)$ diverges if for any $x \in \Bbb R$ there is a neighbourhood $V$ ( or $\epsilon \gt 0$) such that given any $n \in \Bbb N$ there is a corresponding natural number $m_n \ge n$ such that $a_{m_n} \not \in V$ (or $|a_{m_n} - x| \ge \epsilon$ ). This in essence is the negation of the definition of a convergent sequence. But if you are unhappy in general with proofs by contradiction because you are not a fan of the "law of the excluded middle" then my friend you are going to have trouble moving forward.
Let's get to your question.
For any $x \in \Bbb R$ there exists $\epsilon = \frac c 3 \gt 0$ such that given any $n \in \Bbb N$ there is a corresponding natural number $m_n$ fulfilling the prescribed conditions. This must be proven. For any $n$ either $a_n \in \{y \in \Bbb R \ | \ |y - x| \lt \frac c 3\}$ or $a_n \not \in \{y \in \Bbb R \ | \ |y - x| \lt \frac c 3\}$. In the former case $m_n = n + 1$ since $ |a_{n + 1} - x | \ge |a_{n + 1} - a_n| - |x - a_n| \ge c - \frac c 3 = \frac c 2 \gt \frac c 3 $. In the latter trivial case $m_n = n$. This proves the sequence diverges.
There is an easier way though. You may consider the Cauchy Criterion for Convergence. This example is clearly in violation of it.