Proof. By Proposition 4.20.5, there exists an exhaustion $\seq{U_n}\subset 2^{X}$ of $X$ by precompact open sets. Denote $U_{0} = \emptyset$. For each $n \in \natp$, let $V_{n} = U_{n+1}\setminus \ol{U_{n-1}}$.

Let $x \in X$, then there exists $n \in \natp$ such that $x \in U_{n} \setminus U_{n-1}$. In which case, if $n > 1$, then $x \in U_{n}\setminus \ol{U_{n - 2}}= V_{n-1}$. If $n = 1$, then $x \in U_{2}= V_{1}$. Thus $\seq{V_n}$ is an open cover of $X$. In addition, for any $m, n \in \natp$ with $m \le n$, $V_{m} \cap V_{n} \ne \emptyset$ implies that $n - m < 2$, so $\seq{V_n}$ is locally finite. By (2) of Proposition 4.20.8, $X$ is paracompact.$\square$