I am trying to prove that if $K$ is a (non-empty) unbounded subset of a metric space $(X, d)$, then $K$ contains a non-cauchy sequence. Is the following correct?
If $K$ is unbounded, then, for any $M>0$, there exists $x,y\in K$ s.t. $d(x,y)>M$. As such, we may construct the following sequence, in $K$. Let $a_{1}, a_{2}$ be chosen s.t. $d(a_{1}, a_{2})>1$. Generally, for each $n \in \{1,3,5,...\}$, let $a_{n}, a_{n+1}$ be chosen such that $d(a_{n}, a_{n+1})>n$. This completes the construction of $\{a_{n}\}$, which we now prove does not satisfy the Cauchy Criterion.
Consider some arbitrary $\epsilon>0$, and $N \in\mathbb{N}$. Choose $j\in \{1, 3, 5,...\}$ such that $j>\max\{N, \epsilon\}$. Then, set $n=j, m=j+1$. We have that $d(a_{n}, a_{m})=n>\epsilon$, where $m,n>N$. Hence, $\{a_{n}\}_{n\in\mathbb{N}}$ is not Cauchy.
You should exhibit some $\varepsilon>0$ such that, given any $N$, there are $m,n>N$ with $d(a_m,a_n)\ge\varepsilon$.
Your idea is good, but requires a bit of extra work.
You can take $\varepsilon=1/2$. Given $N$, by construction $d(a_{2N+1},a_{2N+2})>2N+1>1/2$, where $2N+1>N$ and $2N+2>N$.