I am reading a famous book by Kolmogorov and Fomin (4th Edition, translated from Russian to Japanese).
Fact:
Let $l^2:=\{x=(x_1,x_2,\dots,x_n,\dots)|x_1,x_2,\dots,x_n,\dots\in\mathbb{R},\sum_{k=1}^\infty x_k^2<\infty\}$.
Let $\rho(x,y):=\sqrt{\sum_{k=1}^\infty (y_k-x_k)^2}$ for any $x,y\in l^2$.
Then $(l^2,\rho)$ is a metric space.
The authors wrote the following proposition holds without a proof:
Proposition:
Let $A:=\{x=(x_1,x_2,\dots,x_n,\dots)|x_1,x_2,\dots,x_n,\dots\in\mathbb{Q}, x_i = 0 \text{ for all }i\geq n_0 \text{ for some }n_0\in\{1,2,\dots,\}\}$.
Then $\overline{A}=l^2$.
Since $A$ is countable, $l^2$ is separable.
Is my proof of this proposition ok?
Proof:
Let $x\in l^2$.
Let $S:=\sum_{k=1}^\infty x_k^2$.
For any positive real number $\epsilon$, there is $N_0\in\{1,2,\dots,\}$ such that $S - \sum_{k=1}^{N_0} x_k^2<\frac{\epsilon}{2}$.
So $x_{N_0+1}^2+x_{N_0+2}^2+\dots<\frac{\epsilon}{2}$.
For each $k\in\{1,2,\dots,N_0\}$, there exists a rational number $y_k$ such that $|y_k-x_k|<\sqrt{\frac{\epsilon}{2N_0}}$.
So, $(y_1-x_1)^2+\dots+(y_{N_0}-x_{N_0})^2<\frac{\epsilon}{2}$.
So, $(y_1-x_1)^2+\dots+(y_{N_0}-x_{N_0})^2+(0-x_{N_0+1})^2+(0-x_{N_0+2})^2+\dots<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.
$y=(y_1,y_2,\dots,y_{N_0},0,0,\dots)\in A$.
So, $x\in \overline{A}$.
So, $l^2=\overline{A}$.
Since $A$ is countable, $l^2$ is separable.
The following proposition is true since $A\subset B$ and $B$ is also countable.
Is my proof of the following proposition ok?
Proposition
Let $B:=\{x=(x_1,x_2,\dots,x_n,\dots)|x_1,x_2,\dots,x_n,\dots\in\mathbb{Q}\}$.
Then $\overline{B}=l^2$.
Since $B$ is countable, $l^2$ is separable.
Proof:
Let $x\in l^2$.
Let $\epsilon$ be any positive real number.
For each $k\in\{1,2,\dots\}$, there is a rational number $y_k$ such that $|y_k-x_k|<\sqrt{\frac{\epsilon}{2^k}}$.
Then, $\sum_{k=1}^\infty (y_k-x_k)^2 < \sum_{k=1}^\infty \frac{\epsilon}{2^k}=\epsilon$.
So, $x\in\overline{B}$.
So, $l^2=\overline{B}$.
Since $B$ is countable, $l^2$ is separable.
Stick to the first proof, which is OK. The dense set $A$ there is indeed countable (but you should show that too, maybe...). Your second set $B$ is just $\Bbb Q^{\Bbb N}$ which has size $|\Bbb R|$ and is not countable. Of course it's dense as it contains $A$, so that needs no new proof anyway.