Say that we have a metric space $X$ and that $Y= \{y_1, y_2,\ldots\}$ is a countable collection of points in $X$ such that for any two points in $Y$, we have $d(y_n, y_m) \geq1$, i.e. the distance between those points is $\geq1$.
We want to show that $Y$ is closed in $X$ and that when equipped with the subspace topology, $Y$ is not compact. Couldn't we go about using contradiction to show the first claim? I'm thinking we could assume $Y$ is not closed in $X$, which would mean that there exists an accumulation point, say $x$ of $Y$, such that $x \notin Y$. How would we go from there? Not exactly sure how to tackle the second part of the question. I guess once the second part is shown, we will in fact have that $X$ is not compact, since every closed subspace of a compact space is compact.