Heine-Borel Theorem/Euclidean Space/Necessary Condition/Proof 1

Theorem
For any natural number $n \ge 1$, a subspace $S$ of the Euclidean space $\R^n$ is closed and bounded iff it is compact.

Necessary Condition
Let $S \subseteq \R^n$ be closed and bounded.

Since $S$ is bounded, $S \subseteq \left[{a \,.\,.\, b}\right]^n = B$ for some $a, b \in \R$.

By the Special Case of the Heine-Borel Theorem and by Topological Product of Compact Spaces, it follows that $B$ is compact in the Tychonoff topology on $\R^n$.

From Euclidean Topology is Tychonoff Topology, it follows that $B$ is compact in the usual Euclidean topology induced by the Euclidean metric.

By the Corollary to Closed Set in Topological Subspace, we have that $S$ is closed in $B$.

By Closed Subspace of Compact Space is Compact, $S$ is compact.

Sufficient Condition
Let $S \subseteq \R^n$ be compact.

From Compact Subspace of Metric Space is Bounded, it follows that $S$ is bounded.

From Metric Space is Hausdorff, it follows that $\R^n$ is a Hausdorff space.

Then Compact Subspace of Hausdorff Space is Closed shows that $S$ is closed.