Suppose that $X$ has a countable basis. Then: (a) Every open covering of $X$ contains a countable subcollection covering $X$. (b) There exists a countable subset of $X$ that is dense in $X$.
Rephrasing this theorem to my taste:
Suppose $X$ is second countable. Then: (a) Every open cover of $X$ has a countable subcover. (b) $\exists D\subseteq X$ such that $D$ is countable and $\overline{D}=X$.
My attempt:
(a) let $U=\{ U_\alpha | \alpha \in J\}$ be an open cover of $X$. $\mathcal{B}=\{ B_n | n\in \Bbb{N}\}$ be countable basis of $\mathcal{T}_X$. $\bigcup_{\alpha \in J} U_\alpha =X$. $\forall x\in X, \exists \alpha_x \in J$ such that $x\in U_{\alpha_x}$. Since $x\in U_{\alpha_x} \in \mathcal{T}_X$, $\exists n_x \in \Bbb{N}$ such that $x\in B_{n_x}\subseteq U_{\alpha_x}$. Then $\{ B_{n_x} | x\in X\}$ is countable, since $\{ B_{n_x} | x\in X\} \subseteq \{ B_n | n\in \Bbb{N}\} =\mathcal{B}$, and $ \bigcup_{x\in X} B_{n_x} =X$. Since $| \{ B_{\alpha_x} | x\in X\}|= | \{ U_{\alpha_x} | x\in X\}|$ and $\{ B_{n_x} | x\in X\}$ is countable, $\{ U_{\alpha_x} | x\in X\}\subseteq U$ is countable. $B_{n_x} \subseteq U_{\alpha_x}$, $\forall x\in X$ $\Rightarrow$ $\bigcup_{x \in X} B_{n_x} =X \subseteq \bigcup_{x \in X} U_{\alpha_x}$. So $\bigcup_{x \in X} U_{\alpha_x} =X$. Thus $\exists \{ U_{\alpha_x}| x\in X\}$ countable subcover of $U$. Is this proof correct? I find Munkres’ proof little bit vague especially, “For each positive integer $n$ for which it is possible, choose an element $A_n$ of $\mathscr{A}$ containing the basis element $B_n$“ sentence.
(b) $\forall n\in \Bbb{N}, \exists x_n \in B_n$, since $B_n \neq \emptyset$ $\forall n\in \Bbb{N}$. Let $D=\{ x_n |n\in \Bbb{N}\}\subseteq X$. Claim: $\overline{D}=X$. Proof: $\overline{D}\subseteq X$ is trivial. Let $x\in X$ and $V\in \mathcal{N}_x$. Since $x\in V\in \mathcal{T}_X$, $\exists m\in \Bbb{N}$ such that $x\in B_m \subseteq V$. So $x_m\in B_m \cap D \subseteq V\cap D$. Thus $V\cap D\neq \emptyset$. Hence $x\in \overline{D}$. Our desired result. Is this proof correct?
Que: (1) $D$ may be finite(still countable by Munkres definition of countable). (2) Is there any other way to construct set $D$?