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

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

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 $X\setminus U_i\mathrm{\ne }\mathrm{\varnothing }$ for every $i\le m$.

For ${\alpha }_0,\dots ,{\alpha }_m$ and ${\beta }_0,\dots ,{\beta }_m$ real one has ${\alpha }_i\le {\beta }_i+\mathrm{\mid }{\alpha }_i-{\beta }_i\mathrm{\mid }\le {\max }_j{\beta }_j+{\max }_j\mathrm{\mid }{\alpha }_j-{\beta }_j\mathrm{\mid }$ for every $i$, so ${\max }_i{\alpha }_i\le {\max }_j{\beta }_j+{\max }_j\mathrm{\mid }{\alpha }_j-{\beta }_j\mathrm{\mid }$, and by symmetry $\mathrm{\mid }{\max }_i{\alpha }_i-{\max }_i{\beta }_i\mathrm{\mid }\le {\max }_i\mathrm{\mid }{\alpha }_i-{\beta }_i\mathrm{\mid }$.

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 $\mathrm{\mid }g(x)-g(y)\mathrm{\mid }\le d(x,y)$, 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 $\mathrm{diam}(A)<\delta$ 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 $d(a,y)\le \mathrm{diam}(A)<\delta \le d(a,X\setminus U_i)$, so $y\mathrm{\notin }X\setminus U_i$, 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 $\mathcal{U}$.