Proposition 13

Let X be a topological space which has a countable base $(U_n)$; then for each open covering $(V_i)_{i \in I}$ of X there exists a countable subset J of I such that $(V_i)_{i \in J}$ is a covering of X.

Let H be the subset of N consisting of indices n such that U_n is contained in at least one of the V_i; the sequence (U_n)_{n \in H} is a covering of X, because every point x \in X belongs to some V_i, and since V_i is open, there is an index n such that x \in U_n \subset V_i. Hence there is a mapping $\psi$ of H into I such that U_n \subset V_{\psi(n)} for each n \in H; taking J = $\psi(H)$, which is countable, the proposition is proved.