We say that $K \subset X$ is relatively compact, if for every sequence $\{x_n\}_{n \in \mathbb{N}} \subset K$ has a Cauchy sub-sequence.
We say that $N_\varepsilon$ is an $\varepsilon\text{-net}$ of $A \subset X$ if for every $a \in A$ there is $y \in N_\varepsilon$ such that $d(a, y) < \varepsilon$.
Now, let $X$ be a metric space. I want to show that if $K \subset X$ is relatively compact, then for every $\varepsilon > 0$ there is a finite $\varepsilon\text{-net}$ for $K$.
My problem here is that $X$ is not necessarily complete, and that means that $\overline K$ is not necessarily compact.
$\overline K$ not being compact makes it harder for me to build the finite $\varepsilon\text{-net}$ for $K$.
I thought maybe covering $K$ with open balls of radius $\varepsilon$ but I am not sure what to do from there, or how the fact that each sequence in $K$ has a Cauchy sub-sequence helps here.
Help would be appreciated.
HINT: Suppose that $\epsilon>0$, and there is no finite $\epsilon$-net in $K$; then you can recursively choose an infinite sequence whose points are all at least $\epsilon$ away from one another.