Every Cauchy sequence in a metric space is bounded

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, $$d(p,x_n)\le d(p,x_m)+d(x_m, x_n) <k+\epsilon.$$

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).

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.