Here is Prob. 5 (a), Sec. 30, in the book Topology by James R. Munkres, 2nd edition:
Show that every metrizable space with a countable dense subset has a countable basis.
My Attempt:
Let $X$ be a metrizable topological space with a countable subset $S$ of $X$ such that $S$ is dense in $X$. Then there exists a metric $d$ on $X$ such that the metric topology on $X$ induced by $d$ is the same as the topology of $X$; also we have a countable set $S$ such that $S \subset X$ and $\overline{S} = X$.
Let $$ \mathscr{B} := \left\{ B_d \left(s, \frac{1}{n} \right) \, \vert \, s \in S, n = 1, 2, 3, \ldots \right\}. \tag{Definition 0} $$ This collection $\mathscr{B}$ is countable since it is indexed by the countable set $S \times \{ 1, 2, 3, \ldots \}$. We show that $\mathscr{B}$ is a basis for the topology of $X$.
Let $U$ be an open set of $X$, and let $p$ be any point of $U$. Then there exists an open ball $B_d (x, \epsilon)$, where $x \in X$ and $\epsilon$ is a positive real nuumber, such that $$ p \in B_d(x, \epsilon) \subset U, \tag{0} $$ since the collection of all open balls in the metric space $(X, d)$ constitutes a basis for the metric topology induced by the metric $d$ on $X$, which is the same as the topology of $X$.
As $p \in B_d (x, \epsilon)$, so we can find a positive real number $\delta$ such that $$ B_d(p, \delta) \subset B_d (x, \epsilon ), $$ which together with (0) above implies $$ B_d (p, \delta) \subset U. \tag{1} $$
Let $n_\delta$ be any natural number such that $$ n_\delta > \frac{2}{\delta}. $$ Then we have $$ \frac{1}{n_\delta} < \frac{\delta}{2}. \tag{2} $$
Now as $p \in X$ and $\overline{S} = X$, so $p \in \overline{S}$ and hence every open ball containing $p$ --- which is a basis neighborhood of $p$ --- must intersect $S$; in particular the open ball $B_d \left( p, \frac{1}{n_\delta} \right)$ intersects $S$, that is, there exists a point $s_0 \in S$ such that $s_0 \in B_d \left( p, \frac{1}{n_\delta} \right)$, which in turn implies that $$ d \left( s_0, p \right) < \frac{1}{n_\delta}. \tag{3} $$ From (3) we can also conclude that $p \in B_d \left( s_0, \frac{1}{n_\delta} \right)$. We show that $$ B_d \left( s_0, \frac{1}{n_\delta} \right) \subset U. \tag{4} $$
Now let $x \in B_d \left( s_0, \frac{1}{n_\delta} \right)$. Then $x \in X$ and $$ d \left( x, s_0 \right) < \frac{1}{n_\delta}. \tag{5} $$ Therefore we find that $$ \begin{align} d (x, p) &\leq d \left( x, s_0 \right) + d \left( s_0, p \right) \\ & \qquad \qquad \mbox{[ using the triangle inequality ]} \\ &< \frac{1}{n_\delta} + \frac{1}{n_\delta} \qquad \mbox{[ using (3) and (5) above ]} \\ &= \frac{2}{n_\delta} \\ &< \delta, \qquad \mbox{[ using (2) above ]} \end{align} $$ which implies that $$ d (x, p) < \delta, $$ and hence that $x \in B_d(p, \delta)$. But $x$ was an arbitrary point of the open ball $B_d \left( s_0, \frac{1}{n_\delta} \right)$. Thus we can conclude that $$ B_d \left( s_0, \frac{1}{n_\delta} \right) \subset B_d ( p, \delta), $$ which together with (1) above implies that (4) above holds.
Thus for every open set $U$ of $X$ and for every point $p \in U$, there exists a set $B_p := B_d \left( s_0, \frac{1}{n_\delta} \right)$ in the countable collection $\mathscr{B}$ [Refer to (Definition 0) above.] such that $$ p \in B_p \subset U. $$ Hence the collection $\mathscr{B}$ is a countable basis for the topology of $X$.
Is this proof correct in each and every detail? If so, is it clear enough in every bit? Or, are there issues?
I guess it's correct, but it can be streamlined. You want to show that the family $$ \mathscr{B} =\{ B_d(s, 1/n ) \mid s \in S, n \in \mathbb{N} \} $$ is a basis for the topology, which is the same as proving that, for every open set $U$ and any $p\in U$, there exists $V\in\mathscr{B}$ such that $$ p\in V\subseteq U $$ Let $n>0$ be such that $B_d(p,1/n) \subseteq U$. Since $S$ is dense, there exists $x\in S$ such that $x\in B_d(p,\delta)$. For every $m>n$, we know there exists $x\in B_d(p,1/m)\cap S$, by density.
Take $m=4n$; then $d(p,x)<1/(4n)<1/(2n)$, so $p\in B_d(x,1/(2n))$; moreover, for $y\in B_d(x,1/(2n))$ it holds $$ d(p,y)\le d(p,x)+d(x,y) < \dfrac{1}{4n}+\dfrac{1}{2n}=\frac{3}{4n}<\frac{1}{n} $$ Therefore $p\in B_d(x,1/(2n))\subseteq B_d(p,1/n)\subseteq U$.