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?
