Here is Prob. 2, Sec. 30, in the book Topology by James R. Munkres, 2nd edition:
Show that if $X$ has a countable basis $\left\{ B_n \right\}$, then every basis $\mathscr{C}$ for $X$ contains a countable basis for $X$. [Hint: For every pair of indices $n, m$ for which it is possible, choose $C_{n, m} \in \mathscr{C}$ such that $B_n \subset C_{n, m} \subset B_m$.]
My Attempt:
As each set $C \in \mathscr{C}$ is an open set of the topological space $X$ and as $\left\{ B_n \right\}$ is a basis for the topology of $X$, so $C$ is a union of some of the sets $B_n$ and thus there exists a set $B_n$ such that $B_n \subset C$, by Lemma 13.1 in Munkres.
Conversely, as each set $B_m$ is an open set in $X$ and as $\mathscr{C}$ is a basis for $X$, so $B_m$ can be written as a union of some sets from $\mathscr{C}$ and thus there exists a set $C \in \mathscr{C}$ such that $C \subset B_m$, again by Lemma 13.1 in Munkres.
Moreover, the collection $\left\{ C_{n,m} \right\}$ is indexed by a subset $J$ of $\mathbb{N}\times\mathbb{N}$ and so is at most countable.
Finally, let $U$ be any open set of $X$ and let $x \in U$. Then there exists a set $B_m$ such that
$$ x \in B_m \subset U. \tag{1} $$
Next, as $B_m$ is an open set in $X$, as $x \in B_m$, and as $\mathscr{C}$ is a basis for the topology of $X$, so there exists a set $C_x \in \mathscr{C}$ such that
$$ x \in C_x \subset B_m. \tag{2} $$
And, as $C_x$ is an open set of $X$ and as $x \in C_x$, so there exists a set $B_n$ such that
$$ x \in B_n \subset C_x. \tag{3} $$
From (2) and (3) we obtain
$$ B_n \subset C_x \subset B_m, $$
which implies that set $C_{n, m}\in \mathscr{C}$ is well-defined such that
$$ B_n \subset C_{n,m} \subset B_m. \tag{4} $$
Finally, from (3), (4), and (1) above, we obtain
$$ x \in B_n \subset C_{n, m} \subset B_m \subset U, $$
and so
$$ x \in C_{n, m} \subset U. \tag{5} $$
Thus for every open set $U$ in $X$ and for every point $x \in U$, there exists a set $C_{n, m}$ such that (5) above holds.
Hence by Lemma 13.2 in Munkres the countable collection $\left\{ C_{n,m} \right\}$ is indeed a basis for $X$.
Is my proof correct and clear enough in each and every detail? Or, are there errors and confusions in it?
Best Answer
The first two paragraphs are superfluous; just follow the hint:
If, for some $(n,m) \in \Bbb N^2$, the set $\mathscr{C}_{n,m}$ given by $$ \mathscr{C}_{n,m} := \left\{C \in \mathscr{C} | B_n \subseteq C \subseteq B_m \right\}$$ is non-empty, pick $C_{n,m} \in \mathscr{C}_{n,m}$. So the collection we pick is indexed by the set $$ J := \left\{ (n,m) \in \mathbb{N}^2 | \mathscr{C}_{n,m} \neq \emptyset \right\} $$ and this set $J \subseteq \Bbb N^2$, so this collection is (at most) countable.
The claim is that the collection $\mathscr{C}^\prime$ given by $$ \mathscr{C}^\prime := \left\{ C_{n,m} | (n,m) \in J \right\}$$ is a basis.
The rest of the argument is fine: If $x \in O$, $O$ open, and by succesive applications of $\mathscr{B}, \mathscr{C}, \mathscr{B}$ being a basis, we get $N, M \in \mathbb{N}$ and $C_x \in \mathscr{C}$ such that $$ x \in B_N \subseteq C_x \subseteq B_M \subseteq O, $$ and so by definition $\mathscr{C}_{N, M} \neq \emptyset$ and thus $(N, M) \in J$ and hence $C_{N, M} \in \mathscr{C}^\prime$ can be used in the same "squeeze" and $$ x \in C_{N,M} \subseteq O, $$ and as $O$ and $x$ were arbitary, $\mathscr{C}^\prime$ is a (countable) basis for the topology of $X$.
The fact that $\mathscr{B}$ is a basis for $X$ iff $$ \forall O \text{ open }: \forall x \in O: \exists B \in \mathscr{B}: x \in B \subseteq O $$ is just a reformulation of the definition, and is constantly used almost implicitly. I gather that you quote it as lemma 13.1?