Every open cover of a compact metric space has a Lebesgue number: a $\delta > 0$ such that every nonempty subset of diameter less than $\delta$ lies inside a single member of the cover

If $X = \emptyset$ then $\delta := 1$ serves, there being no nonempty subset of $X$ to test; assume from now on $X \ne \emptyset$.

Compactness gives $m \in \mathbb{N}$ and $U_0, \dots, U_m \in \mathcal{U}$ with $X = U_0 \cup \dots \cup U_m$.

If $U_i = X$ for some $i \le m$, then $\delta := 1$ serves again, every nonempty $A \subseteq X$ being contained in that $U_i$; assume from now on that $X \setminus U_i \ne \emptyset$ for every $i \le m$.

For $\alpha_0, \dots, \alpha_m$ and $\beta_0, \dots, \beta_m$ real one has $\alpha_i \le \beta_i + |\alpha_i - \beta_i| \le \max_j \beta_j + \max_j|\alpha_j - \beta_j|$ for every $i$, so $\max_i \alpha_i \le \max_j \beta_j + \max_j |\alpha_j - \beta_j|$, and by symmetry $|\max_i \alpha_i - \max_i \beta_i| \le \max_i |\alpha_i - \beta_i|$.

Define $g : X \to \mathbb{R}$ by $g(x) := \max\{\, d(x, X \setminus U_i) : i \le m \,\}$, a maximum of a nonempty finite set of reals; each $x \mapsto d(x, X\setminus U_i)$ changes by at most $d(x,y)$ between $x$ and $y$, so by step 4.1 $|g(x) - g(y)| \le d(x,y)$, and $g$ is Lipschitz with constant $1$, hence continuous.

$g(x) > 0$ for every $x \in X$: such an $x$ lies in some $U_i$ by step 2.1, openness gives a real $r > 0$ with $B(x,r) \subseteq U_i$, so every $y \in X \setminus U_i$ has $d(x,y) \ge r$, making $r$ a lower bound of $\{d(x,y) : y \in X \setminus U_i\}$ and hence $g(x) \ge d(x, X\setminus U_i) \ge r > 0$.

By the extreme value theorem applied to the nonempty compact $X$ and the continuous $g$, there is $x^{\ast} \in X$ with $g(x^{\ast}) \le g(x)$ for every $x$; put $\delta := g(x^{\ast})$, a real with $\delta > 0$ by step 6.1.

Let $A \subseteq X$ be nonempty with $\operatorname{diam}(A) < \delta$ and fix $a \in A$; then $g(a) \ge \delta$, so some $i \le m$ has $d(a, X \setminus U_i) \ge \delta$, the maximum defining $g(a)$ being one of its members, and the least such $i$ may be taken.

Every $y \in A$ satisfies $d(a,y) \le \operatorname{diam}(A) < \delta \le d(a, X\setminus U_i)$, so $y \notin X \setminus U_i$, since a point of that set would make $d(a, X\setminus U_i) \le d(a,y)$; hence $A \subseteq U_i$ with $U_i \in \mathcal{U}$, and $\delta$ is a Lebesgue number for $\mathcal{U}$.