Exercise 1.2.4 (Disjoint compact sets are separated)

Suppose, for the sake of contradiction, that $d(E,F)=0$. In other words, for each $𝜖>0$ we can find a pair $x\in E$ and $y\in F$ such that $d(x,y)<𝜖$. By the axiom of countable choice, we can build a sequence ${(x)}_{n\in \mathbb{N}}$ and ${(y)}_{n\in \mathbb{N}}$ such that

By definition of compactness, every sequence, in particular ${(x_n)}_{n\in \mathbb{N}}$ must contain a convergent subsequence ${(x_{n_k})}_{k\in \mathbb{N}}\to x_0$. Since $E$ is closed, $x_0\in E$. But, then for each $𝜖>0$ we can find an $k\in \mathbb{N}$ large enough so that such that $d(x_0,x_{n_k})<1∕n_k<𝜖∕2$ whereas we also have $d(x_{n_k},y_{n_k})<𝜖∕2$. Thus, by triangle inequality, a sequence of points ${(y_{n_k})}_{k\in \mathbb{N}}$ in $F$ can be eventually $𝜖$-close to the adherent point $x_0$ of $E$, meaning $x_0$ is an adherent point of $F$. By closedness assumption, $x_0\in F$ - thus, we have arrived at the contradiction to the disjointness assumption.