Answer
Proof given below.
Work Step by Step
Let $\{a_{n}\}$ converge to the limit $L.$
Given a positive value $\epsilon$, we observe $\epsilon/2$, for which, by definition of limits, there exists an $N$ such that
$ m \gt N\displaystyle \quad \Rightarrow \quad |a_{m}-L| \lt \frac{\epsilon}{2}\quad$ and
$n \gt N\displaystyle \quad \Rightarrow \quad |a_{n}-L| \lt \frac{\epsilon}{2}$
$|a_{m}-a_{n}| =|(a_{m}-L)+(L-a_{n})|\leq|(a_{m}-L)|+|(L-a_{n})|$
so, if $m\gt N$ and $n\gt N.$
$|a_{m}-a_{n}| \lt \displaystyle \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$
The practical meaning of this result is that, given an arbitrarily small positive number $\epsilon$, there is an index N after which the distance between any two terms that follow, is less than $\epsilon$.
Proof given below.
