Studying for my prelims has been a revealing experience: half a year ago, I proved this claim in an extraordinarily naïve way thinking that it was clever. Although the proof below is my most recent attempt, which is significantly conciser, history may repeat itself. This discipline has humbled me tremendously.

Claim: If $X$ is a metrizable, Lindelöf, topological space, then $X$ is second countable.

Proof: Let $d$ be a metric that induces the topology of $X$, and let $\mathcal B_n=\left\{B_d\left(x,1/n\right):x\in X\right\}$, where $n\in\mathbb N$. Then $\mathcal B_n$ is an open cover of $X$ and thus has a countable sub-cover $\mathcal V_n$. Let $\mathcal B=\bigcup_{n=1}^\infty\mathcal V_n$, let $x\in B_d\left(y_1,1/m_1\right)\cap B_d\left(y_2,1/m_2\right)$ for some $B_d\left(y_1,1/m_1\right),B_d\left(y_2,1/m_2\right)\in\mathcal B$, and let $m_3\in\mathbb N$ such that$$\frac1m_3\leqslant\frac12\min\left\{\frac1{m_1}-d\left(x,y_1\right),\frac1{m_2}-d\left(x,y_2\right)\right\}.$$ Then there is a $B_d\left(y_3,1/m_3\right)\in\mathcal B$ containing $x$. Let $z\in B_d\left(y_3,1/m_3\right)$. Then\begin{align*}d\left(z,y_1\right)&\leqslant d\left(z,y_3\right)+d\left(y_3,y_1\right)\\&\leqslant d\left(z,y_3\right)+d\left(y_3,x\right)+d\left(x,y_1\right)\\&<\frac1{m_3}+\frac1{m_3}+d\left(x,y_1\right)=\frac2{m_3}+d\left(x,y_1\right)\\&\leqslant\frac1{m_1}-d\left(x,y_1\right)+d\left(x,y_1\right)=\frac1{m_1},\end{align*}which implies that $z\in B_d\left(y_1,1/m_1\right)$, which in turn implies that $B_d\left(y_3,1/m_3\right)\subseteq B_d\left(y_1,1/m_1\right)$. A similar argument shows that $B_d\left(y_3,1/m_3\right)\subseteq B_d\left(y_2,1/m_2\right)$. $\blacksquare$