Proof. (1) $\Rightarrow$ (2): Let $\seq{K_n}\subset 2^{X}$ be compact such that $\bigcup_{n \in \natp}K_{n} = X$, and $U_{0} = \emptyset$.

Assume inductively that $\bracs{U_j}_{0}^{n}$ has been constructed such that:

1. For each $0 \le k \le n$, $U_{k}$ is a precompact open set.

2. For each $0 \le k < n$, $\overline{U_k}\subset U_{k+1}$.

3. For each $1 \le k \le n$, $U_{k} \supset \bigcup_{j = 1}^{k} K_{j}$.

By Lemma 4.20.2, there exists $U_{n+1}\in \cn^{o}(\overline{U_n}\cup K_{n+1})$ precompact. In which case, by (c),

Thus $\bracs{U_j}_{0}^{n+1}$ satisfies (a), (b), and (c), and $\seq{U_n}$ is an exhaustion of $X$ by compact sets.$\square$