Every Cauchy sequence in a metric space is bounded Is the following correct or along the right lines?
Thanks for any help
Question
A sequence $\{x_n\}$ in a metric space is said to be bounded if it is contained in some open ball $B(a,r)$.
Prove that every Cauchy sequence in a metric space is bounded.
Proof 
Put $\{x_n\}\in(X,d)$ s.t. $\forall\epsilon>0\ \exists k\in\mathbb{N}\ s.t. \ d(x_n,x_m)<\epsilon$ whenever $n,m\geq k$.
Now for $x_n\geq k$ we have $x_n\in{B(x_n,\epsilon)}$
Defining the finite sequence $y=\{x_1,x_2,\ldots,x_n\}$.
Take $r=\epsilon+\max\{|\max(y)|,|\min(y)|\}$, and so $x_n\in{B(0,r)}$
 A: Choose $N$ so that $m, n\ge N\implies d(x_m, x_n) < 1.$  Using the the triangle inequality, it is fairly easy to see that $x_n \in B_2(x_N)$ for $n \ge N$.  Consequently, we have all but finitely many of the $x_n$ contained in a bounded subset of the space.  Hence, the sequence is bounded.  
A: You seem to have the gist of it, but some of your details are a bit off.
Firstly, I find your first line a tad confusing. You seem to be defining your Cauchy sequence as a sequence that satisfies some condition based on $\epsilon$. What you should be saying is "given a Cauchy sequence $ \{ x_n \} $, fix $ \epsilon > 0 $ and let $k $ be such that $ d(x_n, x_m) < \epsilon $ whenever $n, m \geq k $". Words are often easier to read than symbols and abbreviations like s.t., $ \exists $ etc. In any case, you do not (or should not) mean to use "$\forall$". For the argument to work, $\epsilon$ needs to be fixed. Can you see why?
As Thomas points out, "for $ x_n \geq k $ we have $x_n \in B(x_n , \epsilon) $" is rather obvious. Maybe you made a typo there, but can you see why it is much more useful to observe that for $ x_n \geq k$, we have $ x_n \in B(x_k, \epsilon) $?
So far we've trapped all the terms after (and including) the $k$-th in a ball of size $\epsilon$ around $x_k$. It will help here if we think graphically. We'd now also like to trap the first $k-1$ terms in a ball around $ x_k $, because then we'd know that all of the terms in the sequence lie within a ball around $x_k$ (which, as you stated, is precisely the definition of boundedness), since we can just take the larger of the two balls. 
Well, thankfully, we only have finitely many terms $x_1, ... x_{k-1}$. So we know that (at least) one of them is furthest away from $x_k$ (and that this distance is finite). Call this distance $d$.
So if $ D = \mathrm{max} \{\epsilon, d \} $, then everything in the sequence must lie inside $ B(x_k, D + 1) $ (or $B( x_k, D + \delta)$ where $\delta > 0 $).
A: I have a qualm  with one point in the writing style which I think is worth elucidating (hopefully elucidating):
When you say:
"Put $\{x_n\}\in(X,d)$ s.t. $\forall\epsilon>0\ \exists k\in\mathbb{N}\ s.t. \ d(x_n,x_m)<\epsilon$ whenever $n,m\geq k$."
all you are stating is that $\{x_n\}$ is a Cauchy sequence. You are not fixing a value of $\epsilon$ here, and the statement does not give you a value of $k$ that you can work with.
The statement is passive in this regard: it states that these $\epsilon$'s and $k$'s exist, but it does not provide  specific instances of them by itself.
You have to explicitly state that there is a $k$ that corresponds to a particular and fixed value of $\epsilon$ that you have selected.
So in between the above quoted line and your next line, you'd need to say something like:
"Let $\epsilon>0$"
Now you have a value of $\epsilon$ declared that you can work with. Using the fact that $\{x_n\}$ is Cauchy, you would next say
"since $\{x_n\}$ is Cauchy, there is a $k$ such that..."
(even more nitpicky here, you'd have to add "and we select this value of $k$"; or say something like "since $\{x_n\}$ is cauchy, we may, and do, pick $k$ such that...")
And then you'd proceed with the rest of your proof (with corrections as gleaned from the other fine answers).
A: The intuition behind what you wrote is undoubtedly right. The actual details, not so good.  We give an argument that is in spirit close to what I perceive as your intuition.
Pick a fixed $\epsilon >0$. (Actually, we can choose $\epsilon=1$, or $47$.)  Then there is an $m$ such that if $n \ge m$, then $d(x_m,x_n)<\epsilon$.  Let $p$ be any point in the space, and let 
$$k=\max_{i\le m} \:d(p, x_i).$$
The maximum exists, since $\{x_i\colon\; i\le m\}$ is a finite set.
We show that for any $n$, $d(p,x_n)<k+\epsilon$. This is obvious for $n\le m$, since then $d(p,x_n)\le k$.  For $n>m$, we have, by the Triangle Inequality,
$$d(p,x_n)\le d(p,x_m)+d(x_m, x_n) <k+\epsilon.$$
