Let $(X, \mathcal{T})$ be a metrizable space. Prove that $X$ Lindelöf implies that $X$ is separable.
My attempt:
Let $d$ be the metric that induces the topology $\mathcal{T}$.
For every $n \geq 1$, consider the open cover: $$\{B_X(x,1/n)\mid x \in X\}$$
Because the space is Lindelöf, this implies that for every $n \geq 1$, there is a countable (or finite) subcover given by $$\mathcal{B}_n:= \{B_X(x,1/n) \mid x \in I_n, |I_n| \leq |\mathbb{N}|\}$$
Now, define $D:= \bigcup_{n\geq 1} I_n$. Let $y \in X$. Let $\epsilon > 0$. Choose $m$ a positive integer so large that $1/m < \epsilon$.
Because $y\in X =\bigcup\mathcal{B}_m$, it follows that there exists $x \in D$ such that $y \in B_X(x,1/m)$, meaning that $d(x,y) < 1/m < \epsilon$, such that $x \in B_X(y, \epsilon) \cap D$. Hence, $y \in \overline{D}$ and it follows that $D$ is a dense countable set in $X$, such that $X$ is separable.
Is this correct?
