Lemma 1

Let $X$ be a topological space such that the constant mapping $X \to P$ is proper. Then $X$ is quasi-compact.

(We shall see a little later on (Theorem 1, Corollary 1) that this property characterizes quasi-compact spaces.)

We may restrict ourselves to the case where $X$ is not empty. Let $\mathfrak{F}$ be a filter on $X$, and $X' = X \cup \{ \omega \}$ the topological space associated with $\mathfrak{F}$ ($§ 6$, no. 5, Example). Let $\Delta$ be the subset of $X \times X'$ consisting of all $(x, x)$ where $x \in X$, and let $F = \overline{\Delta}$ be the closure of $\Delta$ in $X \times X'$. In view of the hypothesis on $X$, the image of $F$ under the projection $X \times X' \to X'$ is closed in $X'$; this image contains $X$ and therefore contains $\omega$, which lies in the closure of $X$; in other words, there is a point $x \in X$ such that $(x, \omega) \in F$. By the definition of the topology of $X \times X'$, this means that, for each neighbourhood $V$ of $x$ in $X$ and each $M \in \mathfrak{F}$, we have $(V \times M) \cap \Delta \neq \emptyset$, i.e. $V \cap M \neq \emptyset$, so that $x$ is a cluster point of the filter $\mathfrak{F}$, and therefore $X$ is quasi-compact.

Q.E.D.