Why sets that aren't closed can't be compact?
Let $p$ be a limit point of the set $Y$ that is not contained in $Y$. Then $(U_n)_{n\in\mathbb{Z}^+}$ with
$$U_n := \{ y \in Y : \lvert y - p\rvert > \frac1n\}$$
forms an open cover of $Y$ which has no finite subcover.
Two equivalences which are true for metric spaces are these:
- $A$ is compact if and only if every sequence of elements from $A$ has a convergent subsequence with limit in $A$.
- $A$ is closed if and only if every convergent sequence of elements from $A$ has its limit in $A$.
Now it is clear why a set which isn't closed is not compact. If $A$ is not closed, then there is some sequence $a_n$ of elements of $A$, whose limit is $a\notin A$. But the uniqueness of limits mean that every subsequence of $a_n$ converges to $a$. Therefore there is no convergent subsequence whose limit lies in $A$.
Since this question is tagged (general-topology), it may be good to note that in a general topological space compact sets need not be closed. A simple counterexample is obtained by taking any finite space with a non-discrete topology. All subsets of such a space are trivially compact (any open cover without repeated sets must be finite), but by assumption not all subsets are closed.