Show that a sequence is convergent

If $S\subset \Bbb R$ where $S$ has at least 2 members and $S$ has a member strictly between any 2 of its members, then $S$ cannot be finite.

We show that if $u,v$ are any members of $A'$ with $u<v$ then there exists a member of $A'$ in $(u,v)$ and conclude that if $A'$ has more than 1 member then $A'$ is infinite.

Suppose $u,v\in A'$ with $u<v.$ Let $r=(v-u)/3.$ We show that for any $n\in \Bbb N$ there exists $n'>n$ such that $$(\bullet)\quad x_{n'}\in [u+r,u+2r].$$ So $\{n'\in \Bbb N: x_{n'}\in [u+r,u+2r]\}$ is infinite, so $A'$ has a member in $[u+r,u+2r],$ which is a subset of $(u,v)$.... Here is how:

Given $n\in \Bbb N,$ take $n_1\ge n$ such that $|x_{m+1}-x_m|<r$ whenever $m\ge n_1.$

Now take $n_2\ge n_1$ such that $|u-x_{n_2}|<r,$ which is possible because $u\in A'.$

And $v\in A'$ so take $n_3>n_2$ such that $|v-x_{n_3}|<r. $

We now have $n\le n_1\le n_2<n_3$ and $x_{n_2}<u+r<u+2r<x_{n_3}.$

Finally let $n'$ be the $least$ $j>n_2$ such that $x_j\ge u+r.$

Obviously $n'> n$ (as $n'>n_2\ge n$).

The main point is that $x_{n'-1}<u+r$ and $n'-1\ge n_1$ so $$u+r\le x_{n'}=x_{n'-1} +(x_{n'}-x_{n'-1})<u+r+|x_{n'}-x_{n'-1}|<u+r+r.$$ So $x_{n'}\in [u+r,u+2r] $ as required in $(\bullet)$ above.