Proposition 1

Let $X$ be a topological space.

(i) If the topology of $X$ admits a countable base, $X$ is a Lindelöf space.

(ii) Let $f : X \to Y$ be a continuous mapping from $X$ into a topological space $Y$ (not necessarily separated). If $X$ is a Lindelöf space, the same is true of the subspace $f(X)$ of $Y$.

(i) Let $\mathscr{R}$ be an open covering of $X$, $\mathscr{B}$ a countable base of the topology of $X$, $\mathscr{B}'$ the set of elements $A \in \mathscr{B}$ such that there exists an element of $\mathscr{R}$ containing $A$; $\mathscr{B}'$ is a countable covering of $X$, since every $x \in X$ is contained in a set $U \in \mathscr{R}$, and there exists then $A \in \mathscr{B}$ such that $x \in A \subset U$. There then exists a mapping $\Phi$ from $\mathscr{B}'$ into $\mathscr{R}$ such that one has $A \subset \Phi(A)$ for every $A \in \mathscr{B}'$; it is clear that $\Phi(\mathscr{B}')$ is a countable open covering of $X$ contained in $\mathscr{R}$.

(ii) Let $(V_\lambda)_{\lambda \in L}$ be a family of open sets in $Y$ forming a covering of $f(X)$. The sets $f^{-1}(V_\lambda)$ then form an open covering of $X$, and there exists consequently a countable subset $J$ of $L$ such that the open sets $f^{-1}(V_\lambda)$ for $\lambda \in J$ form a covering of $X$. But then the sets $V_\lambda$ for $\lambda \in J$ form a covering of $f(X)$, which proves that $f(X)$ is a Lindelöf space.