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 $).