Lemma 2

Let $A, B$ be two disjoint closed subsets of a paracompact space $X$. If for each $x \in A$ there is an open neighbourhood $V_x$ of $x$ and an open neighbourhood $W_x$ of $B$ which do not intersect, then there exists an open neighbourhood $T$ of $A$ and an open neighbourhood $U$ of $B$ which do not intersect.

Assuming the truth of this lemma for the moment, we can apply it to the case where $B$ consists of a single point, because $X$ is Hausdorff, and it shows then that $X$ is regular. We can then apply Lemma 2 again to any two disjoint closed subsets of $X$, and this shows that Axiom $(O'_V)$ is satisfied.

To prove the lemma, consider the open covering of $X$ consisting of $CA$ and the $V_x$, where $x \in A$; let $(T_i)_{i \in I}$ be a locally finite open refinement of this covering. Then, by definition, if $A \cap T_i \neq \emptyset$ there exists $x_i \in A$ such that $T_i \subset V_{x_i}$. Let $T$ be the open set which is the union of the $T_i$ which meet $A$, and let us show that there is an open neighbourhood $U$ of $B$ which does not meet $T$. For each $y \in B$ there is an open neighbourhood $S_y$ of $y$ which meets only a finite number of sets $T_i$; let $J$ be the finite subset of $I$ consisting of those indices $i$ such that $T_i$ meets both $S_y$ and $A$; if we put $U_y = S_y \cap \bigcap_{i \in J} W_{x_i}$, then $U_y$ is an open neighbourhood of $y$ which meets none of the $T_i$ which meet $A$, and hence $U_y \cap T = \emptyset$. Let $U = \bigcup_{y \in B} U_y$; then $U'$ is an open neighbourhood of $B$ which does not meet $T$, and the lemma is proved.

There exist normal spaces which are not paracompact (Exercise 19).