In my understanding, to prove that $\ell_p$ is a separable space, one may take the subspace of rational sequences, which satisfy the following condition:
$$\sum\limits_{k=1}^\infty |x_k|^p\in \mathbb Q\mbox{ with $\{x_k\}$ a rational sequence}$$
Then this space is countable and dense in $\ell_p$.
- Countability:
Since $\mathbb Q$ is countable, there exists an injection $f:\mathbb Q\to \mathbb N$. Thus there exists an injection $g:(\ell_p)_\mathbb{Q}\to \mathbb N$, defined by $g\left[(x_k)_{k=1}^\infty\right]=f\left[\sum\limits_{k=1}^\infty |x_k|^p\right]\in \mathbb N$.
- Density:
Let $(y_i)_{i=1}^\infty\in \ell_p$, then, since $|y_i|\in \mathbb R$ for all $i\in\mathbb{N}$, there is a sequence $(x_k)_{k=1}^\infty\in (\ell_p)_\mathbb{Q}$ (with $|x_k|\in\mathbb Q$) such that
$$\forall \varepsilon >0, |y_i-x_k|^p <\varepsilon, \mbox{ for some $k\in\mathbb{N}$ }$$
Do you think this proof is good enough, or am I missing some important point(s)?
The set you describe, let's call it $X,$ is uncountable. Proof: Define $x_n = 1/2^{n/p}, n\in \mathbb N.$ Then $(x_n)\in X.$ Note that $x_m\ne x_n$ if $m\ne n.$ It follows that if $\sigma_1, \sigma_2$ are distinct bijections of $\mathbb N$ to $\mathbb N,$ then $x_{\sigma_1(n)}, x_{\sigma_2(n)}$ are distinct sequences in $X.$ Since the set of such bijections is uncountable, so is $X.$
I'm not sure why you're going to all this trouble. An easy countable dense subset of $l^p$ is the set of rational sequences in $l^p$ that are $0$ except for finitely many terms.
Added later: @DanielWainfleet pointed out that in my first answer I neglected the requirement that the sequences in $X$ should have all rational terms. To achieve that, inductively choose distinct rational $x_1,x_2,\dots$ such that
$$\frac{1-(x_1^p + \cdots + x_n^p)}{2} < x_{n+1}^p < 1-(x_1^p + \cdots + x_n^p).$$
for all $n.$ Then $\sum_{n=1}^{\infty}x_n^p = 1.$ The rest of the argument is the same.