I am a beginner in metric space course. Recently I have learnt the terms second countability and Lindelöfness of a space(metric/topological).Now I have proved that second countability implies Lindelöfness in a space. I want to get it verified whether the proof is a correct one:
Suppose $(B_n)$ is a countable base of $X$.Let us take an open cover $\{G_\alpha:\alpha \in \lambda\}$ of $X$.for each $x\in X$ there exists a $G_\alpha$ such that $x\in G_\alpha$. Choose one such set and call it $G_{\alpha_x}$.Then there is an element $B$ of $(B_n)$such that $x \in B \subset G_{\alpha_x}$. Choose one such $B$ and call it $B_x$.Consider the set $\{B_x:x \in X\}$ which is a subset of the countable base and hence countable. Each element of this set is contained in some element of $\scr G_0$$=\{G_{\alpha_x}:x\in X\}$.For each element of $\{B_x\}_{x\in X}$,choose one such set from $\scr G_0$,call this collection $\scr G$ and this is countable and it covers $X$.So $\scr G$ is the required open cover of $X$.So $X$ is Lindelöf.
You proof is fine, but you can also write it a different way:
Let $(B_n)_{n \in \omega}$ be a countable base for $X$ and for the cover $\mathcal{G}=\{G_i: i \in I\}$ define
$$N=\{n \in \omega: \exists i_n \in I: B_n \subseteq G_{i_n}\}$$
which is a well-defined countable set. My claim is that $\mathcal{G}':=\{G_{i_n}: n \in N\}$ is a subcover of $\mathcal{G}$: if $x \in X$ we can find $G_j$ in $\mathcal{G}$ such that $x \in G_j$ and a basic $B_m$ such that $x \in B_m \subseteq G_j$. Then $j$ witnesses that by definition $m \in N$ and then $G_{i_m} \in \mathcal{G}'$, which by definition contains $B_m$, contains $x$ too. QED.
This write-up only uses countable choice (one choice of $i_n$ per $n \in N$). Minor point, admittedly.