Q:

Prob. 4, Sec. 30, in Munkres' TOPOLOGY, 2nd ed: Every compact metrizable space has a countable basis.

Here is Prob. 4, Sec. 30, in the book Topology by James R. Munkres, 2nd edition:

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: 

Let $X$ be a compact metrizable topological space; let $d$ be a metric on $X$ such that the metric topology determined by $d$ is the same as the topology of $X$.
Then the collection of all the open balls is a basis for the topology of $X$.
Let $n \in \mathbb{N}$, and let 
  $$ \mathscr{A}_n \colon= \left\{ \, B_d \left( x, \frac{1}{n} \right) \, \colon \, x \in X \, \right\}. \tag{Definition 0} $$
  Then $\mathscr{A}_n$ is an open covering of the compact space $X$; so $\mathscr{A}_n$ has a finite subcollection 
  $$ \mathscr{B}_n \colon= \left\{ \, B_d\left( x_{n, 1}, \frac{1}{n} \right), \ldots, B_d\left( x_{n, N_n}, \frac{1}{n} \right) \, \right\} \tag{Definition 1} $$
  which also covers $X$, where $x_{n, 1}, \ldots, x_{n, N_n}$ are some points in $X$.
Let us now put
  $$ \mathscr{B} \colon= \bigcup_{n \in \mathbb{N} } \mathscr{B}_n. \tag{Definition 2} $$
This $\mathscr{B}$ is a union of countably many collections each consisting of finitely many sets; so the collection $\mathscr{B}$ also consists of countably many sets.
We now show that $\mathscr{B}$ is a basis for the toplogy of $X$. 
We will apply Lemma 13.2 in Munkres. 
Let $p \in X$, and let $U$ be any open set of $X$ such that $p \in U$. Then there exists a real number $\epsilon > 0$ and a point $x \in X$ such that 
  $$ p \in B_d (x, \epsilon) \subset U. $$
  We can then find a real number $\delta > 0$ such that
  $$ B_d(p, \delta) \subset B_d(x, \epsilon). $$
  Therefore we obtain
  $$ B_d(p, \delta) \subset U. \tag{0} $$

What next? How to proceed from here to show that there exists a ball in $\mathscr{B}$ above that contains point $p$ and is contained in set $U$?
PS:

Let $n_\delta$ be any natural number such that $$ \frac{1}{n_\delta} < \frac{\delta}{2}. \tag{1} $$
Since $\mathscr{B}_{n_\delta}$ is a covering of $X$, so $p \in B_d \left( x_{n_\delta, i}, \frac{1}{n_\delta} \right)$. [Refer to (Definition 0) and (Definition 1) above.] Then we have 
  $$ d \left( p, x_{n_\delta, i} \right) < \frac{1}{n_\delta}. \tag{2} $$
Now let $y \in B_d \left( x_{n_\delta, i}, \frac{1}{n_\delta} \right)$. Then $y \in X$ such that 
  $$ d \left( y, x_{n_\delta, i} \right) < \frac{1}{n_\delta}. \tag{3} $$
  So from (2), (3), and (1) above, we find that
  $$
d (y, p) \leq d \left( y, x_{n_\delta, i} \right) + d \left( x_{n_\delta, i}, p \right) < \frac{1}{n_\delta} + \frac{1}{n_\delta} < \frac{\delta}{2} + \frac{\delta}{2} = \delta,
$$
  which shows that $y \in B_d(p, \delta)$, and thus it follows that
  $$ B_d \left( x_{n_\delta, i}, \frac{1}{n_\delta} \right) \subset B_d(p, \delta), $$
  which together with (0) and (2) above implies that
  $$ p \in B_d \left( x_{n_\delta, i}, \frac{1}{n_\delta} \right) \subset U. \tag{4} $$
  And, also 
  $$ B_d \left( x_{n_\delta, i}, \frac{1}{n_\delta} \right) \in \mathscr{B}. $$
  [Refer to (Definition 2) above.] Let us put
  $$ B_p \colon= B_d \left( x_{n_\delta, i}, \frac{1}{n_\delta} \right). \tag{Definition 3} $$
Thus from (4) we have shown that, for any open set $U$ of $X$ and for any point $p \in U$, there exists a set $B \in \mathscr{B}$ [correction: a set $B_p \in \mathscr{B}$] such that 
  $$ p \in B_p \subset U. $$
  [Refer to (Definition 3) above.] Moreover, $\mathscr{B}$ is a collection of some open sets of $X$. 
So by Lemma 13.2 in Munkres $\mathscr{B}$, which is a collection consisting of countably many sets, is a basis for the topology of $X$, as required. 
Hence every compact metrizable space is second-countable.

Is this proof complete now? If so, is it correct and clear enough in each and every detail? Or, do issues still remain?

A:

If $U$ is an open set, and $p\in U$, there is an $\epsilon>0$ such that $B_d(p,\epsilon)\subseteq U$. Now choose $n\in\Bbb N$ large enough that $\frac2n<\epsilon$; there is a $B\in\mathscr{B}_n$ such that $p\in B$. Show that $B\subseteq U$.

