proof of Lindelöf theorem

Let $X$ be a second countable topological space, $A\subseteq X$any subset and $𝒰$ an open cover of $A$. Let$ℬ$ be a countable basis for $X$; then ${ℬ}^{\prime }=\{B\cap A:B\in ℬ\}$ is a countable basis of thesubspace topology on A. Then for each $a\in A$ there is some$U_a\in 𝒰$ with $a\in U_a$, and so there is$B_a\in {ℬ}^{\prime }$ such that $a\in B_a\subseteq U_a$.

Then $\{B_a\in {ℬ}^{\prime }:a\in A\}\subseteq ℬ$ isa countable open cover of $A$. For each $B_a$, choose$U_{B_a}\in 𝒰$ such that $B_a\subseteq U_{B_a}$. Then$\{U_{B_a}:a\in A\}$ is a countable subcover of $A$ from$𝒰$.$\mathrm{\square }$