Proof. Let $\fF \subset 2^{X}$ be the tail filter of $\angles{x_\alpha}_{\alpha \in A}$, then by the ultrafilter lemma, there exists an ultrafilter $\fU \subset 2^{X}$ with $\fU \supset \fF$. For each $U \in \fU$, $U \cap F \ne \emptyset$ for all $F \in \fF$. Hence $\angles{x_\alpha}_{\alpha \in A}$ is frequently in $U$ for all $U \in \fU$. By Lemma 5.3.9, $\angles{x_\alpha}_{\alpha \in A}$ admits a subnet $\angles{x_\beta}_{\beta \in B}\subset \angles{x_\alpha}_{\alpha \in A}$ such that for each $U \in \fU$, $\angles{x_\beta}_{\beta \in B}$ is in $U$ eventually. As $\fU$ is an ultrafilter, this subnet is an ultranet.$\square$