Theorem 2 (Borel-Lebesgue)

For a subset of the real line $\mathbf{R}$ to be compact it is necessary and sufficient that it be closed and bounded.

1) The condition is necessary. Let $A$ be a compact subset of $\mathbf{R}$ and let $a$ be a real number $> 0$. The set $A$ is closed (Chapter I, $\S$ 9, no. 3, Proposition 4) and there exists a finite number of points $x_i$ ($1 \leq i \leq n$) of $\mathbf{R}$ such that $A$ is contained in the union of the neighbourhoods $[x_i - a, x_i + a]$ (Chapter I, $\S$ 9, no. 3). Let $b$ be the maximum of the numbers $|x_i|$; then we have $A \subset [-b - a, b + a]$.

2) The condition is sufficient. It is enough to show that every interval $[ -a, +a ]$ ($a > 0$) is compact, and since this interval is a closed subset of a complete uniform space, it is enough to show that, for each $b > 0$, we can cover $[ -a, +a ]$ by a finite number of intervals of the form $[x - b, x + b]$ (Chapter II, $\S$ 4, no. 2, Corollary to Theorem 3). Now, let $n$ be an integer $> 0$ such that $a < nb$; if $x \in [ -a, +a ]$ and if $m$ is the largest integer (positive or negative) such that $mb \leq x$, then we have $-n \leq m \leq n$ and $mb \leq x \leq (m + 1)b$. Hence the $2n + 1$ intervals $[ (k - 1)b, (k + 1)b ]$ ($-n \leq k \leq n$) form a covering of the required type.