Let $a_1,a_2,\ldots,a_n,\ldots$ be a convergent sequence in a metric space $(X,d)$. If the set $S$ consists of the (distinct) points in this sequence, show that $S$ is a bounded set.
My arguments: every convergent sequence is a Cauchy sequence. Every Cauchy sequence is bounded. Any subset of bounded set is bounded. Hence $S$ is bounded.
Why $S$ consists of the (distinct) points - does it matter? Are there any examples when $S$ is not bounded?
On
Every Convergent sequence is bounded and we really do not need Cauchy to help us with the proof.
Let us call the limit $L$
For example if you let $\epsilon =1$, eventually for some N, you have $$ n\ge N \implies |a_n-L|<1.$$
Now the set $$\{ a_1, a_2,..., a_{N-1}\} $$is bounded because it is finite.
Also the set {$a_N$, $a_{N+1}$,....} is bounded because of $$|a_n-L|<1$$
Thus your set is bounded.
Any convergent sequence is bounded. Let $a$ be the limit. By definition, there is an $N$ such that for all $n \geq N$, $d(a, a_n) < 1$ for all $n \geq N$. Then let $$ D = \max\{d(a, a_1), \ldots, d(a, a_{N - 1})\}, $$ which is finite since it is the maximum of a finite set of numbers. Then each $a_k$ is in the ball $B(a, r)$ where $r = \max\{1, 2D\}$. The idea is that we can bound all the large terms by their being close to the limit $a$, and we can bound the earlier terms simply because there are only finitely many of them.