Here is Prob. 5 (b), Sec. 30, in the book Topology by James R. Munkres, 2nd edition:
Show that every metrizable Lindelof space has a countable basis.
My Attempt:
Let $X$ be a metrizable Lindelof space; let $d$ be a metric on $X$ such that the metric topology induced by $d$ is the same as the topology of $X$.
Then for each $n \in \mathbb{N}$, the collection $$\mathscr{A}_n := \left\{ B_d(x, 1/n ) | x \in X \right\} \tag{Definition 0} $$ is an open covering of $X$, and as $X$ is Lindelof, so every open covering of $X$ has a countable subcollection that also covers $X$; in particular, the open covering $\mathscr{A}_n$ given in (Definition 0) above too has a countable subcollection that also covers $X$; let $$ \mathscr{B}_n := \left\{ B_d \left( x_m, 1/n \right) | m \in \mathbb{N}, x_m \in X \right\} \tag{Definition 1} $$ be one such countable subcollection. Now let us put $$ \mathscr{B} := \bigcup_{n \in \mathbb{N} } \mathscr{B}_n. \tag{Definition 2} $$ This collection $\mathscr{B}$, being a countable union of countable collections, is also countable. Moreover, $\mathscr{B}$ is a collection comprised of some open sets of $X$.
We show that the countable collection $\mathscr{B}$ in (Definition 2) is a basis for the topology of $X$.
Let $U$ be any open set of $X$, and let $p$ be any point of $U$. Then there exists an open ball $B_d(x, \epsilon)$, where $x$ is some point of $X$ and $\epsilon$ is some positive real number, such that $$ p \in B_d(x, \epsilon) \subset U, \tag{0} $$ because the collection of all open balls centered at points of $X$ constitutes a basis for the topology of $X$, which is the same as the metric topology induced by the metric $d$ on $X$. And, as $p \in B_d(x, \epsilon)$, so we can find a positive real number $\delta$ such that $$ B_d (p, \delta) \subset B_d(x, \epsilon). $$ The last inclusion together with (0) above yields $$ B_d(p, \delta) \subset U. \tag{1} $$
Let $n_\delta$ be any natural number greater than $4/\delta$. Then we obtain $$ \frac{1}{n_\delta} < \frac{\delta}{4}. \tag{2} $$
Now as the countable collection $\mathscr{B}_{n_\delta}$ in (Definition 1) above constitutes a covering (in fact an open covering) of $X$, so there exists a natural number $m_p$ such that $$ p \in B_d \left( x_{m_p}, \frac{1}{n_\delta} \right), $$ [Here $x_{m_p} \in X$ of course.] which implies that $$ d \left( p, x_{m_p} \right) < \frac{1}{n_\delta}. \tag{3} $$
Let $y$ be any point of $B_d \left( x_{m_p}, \frac{1}{n_\delta} \right)$. Then by definition $y \in X$ such that $$ d \left( y, x_{m_p} \right) < \frac{1}{n_\delta}. \tag{4} $$ Then we obtain $$ \begin{align} d (y, p) &\leq d\left( y, x_{m_p} \right) + d \left( x_{m_p}, p \right) \\ & \qquad \mbox{[ using the triangle inequality ]} \\ &< \frac{1}{n_\delta} + \frac{1}{n_\delta} \qquad \mbox{[ using (3) and (4) above ]} \\ &< \frac{\delta}{4} + \frac{\delta}{4} \qquad \mbox{[ using (2) above ]} \\ &< \delta, \end{align} $$ and hence $d (y, p) < \delta$, which implies that $y \in B_d(p, \delta)$ and since by (1) above $B_d(p, \delta) \subset U$, this $y \in U$ also. But $y$ was an arbitrary point of $B_d \left( x_{m_p}, \frac{1}{n_\delta} \right)$; thus we can conclude that $$ B_d \left( x_{m_p}, \frac{1}{n_\delta} \right) \subset U. \tag{5} $$ Moreover, $B_d \left( x_{m_p}, \frac{1}{n_\delta} \right) \in \mathscr{B}$ in (Definition 2) above.
Thus we have shown that, the collection $\mathscr{B}$ of (Definition 2) above is a collection of some open sets of $X$ such that, for every open set $U$ of $X$ and for evey point $p \in U$, there exists a set $B_p := B_d \left( x_{m_p}, \frac{1}{n_\delta} \right)$ in the collection $\mathscr{B}$ such that $$ p \in B_p \subset U. $$ Refer to (5) above.
Hence the countable collection $\mathscr{B}$ of (Definition 2) above is a countable basis for the topology of $X$, as required.
Is this proof correct and accurate in each and every detail? If so, is my presentation clear enough too? Or else, are there any problems of accuracy or clarity anywhere?
Simpler proof for the same $\mathcal{B}$:
Let $B(p, r)$ be any open ball, $p \in X$ $r>0$. Let $n$ be such that $\frac1n < \frac r2$. There must be some $B(x_m, \frac1n) \in \mathcal{B}_n\subseteq \mathcal B$ that contains $p$. So $d(p,x_m) < \frac 1n$ and so $p \in B(x_m, \frac1n) \subseteq B(p, r)$ (as $x' \in B(x_m, \frac1n)$ implies $d(p, x') \le d(p, x_m) + d(x_m, x') < \frac1n + \frac1n < r$). It follows that $\mathcal{B}$ is a base for $(X,d)$.