for a compact space $X$ and an open cover $\{ U_\alpha \}$, there is a radius $r > 0$ such that any ball of such a radius will be in some open cover: For all $x \in X$, for all such balls $B(x, r)$, there exists a $U_x \in \{ U_\alpha \}$ such that $B(x, r) \subseteq U_x$. Intuitively, pick a point $x$. for each open $U$, we have a ball $B(x, r)$ that sits inside it since $U$ is open. Find the largest such radius, we can do so since $\{x\}$ is the closed subset of a compact set. This gives us a function $f$ that maps a point $x$ to the largest radius of ball that can fit in some open cover around it. This function $f$ is a continuous function (why?) on a compact set, and thus has a minimum. So, for all points $x \in X$, if you give a ball of radius $\min f$, I can find some open cover around it.

§ Lebesgue number lemma, Version 2:

for a compact space $X$ and an open cover $\{ U_\alpha \}$, there is a diameter $d > 0$ such that set of smaller radius will be in some open cover: For all $x \in X$, for all opens $x \in O$ such that $diam(O) < d$, there exists a $U_x \in \{ U_\alpha \}$ such that $O \subseteq U_x$.

If we can find radius $B(x, r)$ that satisfies this, then if we are given a set of diameter less than $2r$, there will be a ball $B(x, r)$ that contains the set $O$ of diameter at most $d = 2r$, and this ball will be contained in some $U_x$. So we will have the containments $O \subseteq B(x, r) \subseteq U_x$.

§ Lebesgue number lemma, proof from Hatcher

TODO