Lemma 1

If $x$ is any point of $X$, there is a distinguished pair $(W, U)$ such that $x \in W$.

The inverse image $\overline{p}^{-1}(p(x))$ is discrete and therefore there is a neighbourhood of $x$ in $X$ all of whose points $x'$ other than $x$ have an image $p(x') \neq p(x)$; since $X$ satisfies $(O_{III})$, there is a closed neighbourhood $V$ of $x$ with this property, and we may assume also that $V$ is contained in a set of $\mathfrak{B}$. Let $F$ be the frontier of $V$ in $X$. By condition (i) of the theorem, $p(F)$ is closed in $Y$; and since $p(F)$ does not contain $p(x)$, there is a set $U \in \mathfrak{B}$ which contains $p(x)$ and does not meet $p(F)$. Let $W$ be the component of $x$ in $\overline{p}^{-1}(U)$; then it is enough to show that $W \subset A$. If this were not so, then $W$ would meet $F$ (no. 1, Proposition 3) and therefore $p(F)$ would meet $U$, contrary to the definition of $U$.