I) Let $b_1,b_2,...,b_n....$ be a Cauchy sequence in a complete metric space $(X,d)$. If $T$ is the set of points in the sequence show that $T$ is a bounded set.
As the space is complete, every Cauchy sequence is convergent and every convergent sequence is bounded. As $T\subseteq\{b_i:i\in\mathbb{N}\}$ then $T$ is bounded.
II) Is I) above still true if we do not insist that $(X,d)$ is complete?
If I) is not true then the sequence is not necessarily convergent in $X$ so it remains to show if every Cauchy sequence is bounded.
I devised the following proof:
As $\{b_i:i\in\mathbb{N}\}$ is a Cauchy sequence then $\forall\epsilon>0\exists m,n>N\in\mathbb{N}$ such that $d(b_n,b_m)<\epsilon$.
$\max_{i\in \mathbb{N}}{d(b_i,b_n)}$ must be smaller than some $L\in\mathbb{R}$, which implies that the $\max_{i,j\in \mathbb{N}}{d(b_i,b_j)}\leqslant \max_{i\in \mathbb{N}}{d(b_i,b_n)}+d(b_n,b_m)<L+\epsilon $, proving the Cauchy sequence is bounded.
Question:
Is my proof right? If not. What are the alternatives?
Thanks in advance!
The proof is not correct. For instance, asserting that your sequence is a Cauchy sequence means that$$(\forall\varepsilon>0)(\exists \in\mathbb{N})(\forall m,n\in\mathbb{N}):m,n\geqslant N\implies d(b_m,b_n)<\varepsilon;\tag1$$that is not what you wrote.
It follows from $(1)$ that there is a natural $N$ such that $m,n\geqslant N\implies d(b_m,b_n)<1$. Therefore, each $b_n$, with $n\geqslant N$, belongs to the ball $B_1(x_N)$. So, the set $\{b_n\,|\,n\geqslant N\}$ is bounded. The set $\{b_1,b_2,\ldots,b_{N-1}\}$ is also bounded, since it is finite. So, your set is bounded, since it is the union of two bounded sets.