Show that every compact metrizable space $X$ has a countable basis. [Hint: Let $\mathscr{A}_n$ be a finite covering of $X$ by $1/n$-balls.]
My attempt:
Approach(1): $B_n =\{ B(x, \frac{1}{n})| x\in X\}$ is an open cover of $X$, $\forall n\in \Bbb{N}$. Since $X$ is compact, $\exists A_n =\{ B(x_{n,i}, \frac{1}{n})| i\in J_{k_n}\}$ finite subcover of $B_n$, $\forall n\in \Bbb{N}$. By elementary set theory, $\bigcup_{n\in \Bbb{N}}A_n = \{ B(x_{n,i},\frac{1}{n})| i\in J_{k_n}, n\in \Bbb{N}\}$. Let $\mathcal{B}= \bigcup_{n\in \Bbb{N}}A_n$. By theorem 2.12 of Baby Rudin, $\mathcal{B}$ is countable. (1) Clearly $\mathcal{B} \subseteq \mathcal{B}_X \subseteq \mathcal{T}_X$, where $\mathcal{B}_X =\{ B(x,r)| x\in X, r\in \Bbb{R}^+\}$. (2) let $U\in \mathcal{T}_X$ and $x\in U$. Since $X$ is metrizable, $\exists \epsilon \gt 0$ such that $B(x,\epsilon)\subseteq U$. By theorem 1.20(a) of Baby Rudin, $\exists n\in \Bbb{N}$ such that $\frac{1}{n}\lt \frac{\epsilon} {2}$. Since $x \in X=\bigcup A_n= \bigcup_{i=1}^{k_n} B(x_{n,i},\frac{1}{n})$, $\exists m\in J_{k_n}$ such that $x\in B(x_{n,m},\frac{1}{n})$. Now we show $B(x_{n,m}, \frac{1}{n}) \subseteq B(x,\epsilon)$. Let $p\in B(x_{n,m}, \frac{1}{n})$. Then $d(x_{n,m}, p)\lt \frac{1}{n}$. $d(p,x)\leq d(p, x_{n,m}) +d(x_{n,m},x) \lt \frac{2}{n} \lt \epsilon$. So $p\in B(x,\epsilon)$. Thus $B(x_{n,m}, \frac{1}{n}) \subseteq B(x,\epsilon)$. Hence $\exists B(x_{n,m}, \frac{1}{n}) \in \mathcal{B}$ such that $x\in B (x_{n,m}, \frac{1}{n}) \subseteq B(x,\epsilon)\subseteq U$. By lemma 13.2, $\mathcal{B}$ is basis of $\mathcal{T}_X$. Is this proof correct?
Proof of Exercise 5, Section 30 of Munkres’ Topology, 5(b) precisely, is generalization of this proof. Note: In proof of exercise 4 section 30, we make use of hint, without hint it is very hard to solve.
Approach(2): Since $X$ is compact, $X$ is lindelof. By Exercise 5, Section 30 of Munkres’ Topology, $X$ is second countable.
Best Answer
Extended comment about a slightly different approach.
Let $(X,d)$ be a metric space. For $q\in\Bbb Q^+$ let $F(q)$ be a $\subset$-maximal family of pairwise-disjoint open balls of radius $q.$ For $\beta\in F(q)$ let $\beta=B_d(c_{q, \beta},q).$ Let $D(q)=\{c_{q,\beta}: \beta\in F(q)\}.$
$(1)$. Suppose some $F(q)$ is uncountable. Then $D(q)$ is uncountable. Let $H=\bigcup \{B_d(x,q):x\in X\setminus\bigcup F(q)\}.$ Let $C=F(q)\cup \{H\}.$ Then $C$ is an uncountable open cover of $X$ and each member of the uncountable set $D(q)$ belongs to exactly one member of $C,$ so $X$ is not Lindelof.
Also $d(x,y)\ge q$ for any unequal $x,y\in D(q),$ so $D(q)$ is an uncountable closed discrete subspace of $X.$
Also, if $B$ is any base (basis) for $X$ then for each $x\in D(q)$ there exists $b(x)\in B$ such that $x\in b(x)\subset B_d(x,q)\in F(q),$ and $\{b(x):x\in D(q)\}$ is an uncountable pairwise-disjoint subset of $B,$ so $B$ is uncountable.
$(2).$ Suppose no $F(q)$ is uncountable. The $\subset$-maximality of each $F(q)$ ensures that $\cup_{q\in \Bbb Q}D(q)$ is dense in $X.$ In general, by a method similar to that in the Q, if $D$ is a dense subset in a metric space $(X,d)$ then $\{B_d(x,r):x\in D\land r\in\Bbb Q^+\}$ is a base (basis) for $X.$
Further remarks: In the study of topological vector-spaces we have "Hamel basis", "Schauder basis", & "Hilbert-space basis" so some of us prefer "base" for topological "basis". A quirk of English is that the plurals of base and basis are both spelled "bases"...... A non-metrizable normal $(T_4)$ space that is not Lindelof can fail to have any closed infinite discrete subspace.