A Euclidean space of fixed finite dimension n also forms a metric space with the Euclidean distance as metric. As a topological space, the same statement holds: a subset is compact if and only if it is closed and bounded.
The theorem makes two assertions. Firstly, that a compact subset of R is closed and bounded. A compact subset of any Hausdorff space is closed. The metric is a continuous function on the compact set, and a continuous function on a compact set is bounded.
The second and major part of the theorem is that a closed bounded subset of R is compact. We may reduce to the case of a closed interval, since a closed subset of a compact space is compact.
One proof in this case follows directly from the definition of compactness is terms of open covers. Consider an open cover Uλ. Let S be the subset of the closed interval [a,b] consisting of all x such that the interval [a,x] has a finite subcover. The set S is non-empty, since a is in S. If b were not in S, consider the supremum s of S, and show that there is another t between s and b which is also in S. This contradiction shows that b is in S, which establishes the result.
A second proof relies on the Bolzano-Weierstrass theorem to show that a closed interval is sequentially compact. This already shows that it is countably compact. But R is separable since the rational numbers Q form a countable dense set, and this applies to any interval as well. Hence countable compactness implies compactness.
Finally we note that a finite product of compact spaces is compact, and a closed bounded subset of Rn is a closed subset of a "closed box", that is, a finite product of closed bounded intervals.