Let me offer a simpler proof. First, let $c_{00}$ denote the subset of $\ell_2$ that consists of sequences with finite support, that is, sequences with finitely many non zero terms. Then your set, let's call it $r$, with rational entries, is a subset of this. I claim that $r$ is dense in $c_{00}$ and $c_{00}$ is in turn dense in $\ell_2$.
Proof Given $\varepsilon >0$ and $x\in c_{00}$, say $x=(x_1,x_2,\ldots,x_n,0,0,0,\ldots)$ pick rationals $r_1,\ldots,r_n$ such that $$|r_i-x_i|^2<2^{-i+1}\varepsilon$$
for $i=1,\ldots,n$. We can do this, for $\Bbb Q$ is dense in $\Bbb R$. Let, $q=(r_1,\ldots,r_n,0,\ldots)\in r$. Then we see $$\lVert x-q\rVert^2 =\sum_{i=1}^n|x_i-r_i|^2<\varepsilon\sum_{i=1}^n2^{-i+1}\leqslant \varepsilon\sum_{i=1}^\infty 2^{-i+1}=\varepsilon$$
This proves $r$ is dense in $c_{00}$. Now, we'll prove $c_{00}$ is dense in $\ell_2$. Pick thus $x\in\ell^2$. By definition, $$\sum_{n\geqslant 1}x_n^2<+\infty$$
where $x=(x_1,x_2,\ldots)$. In particular, given $\varepsilon >0$; there exists $N$ large enough so that $$\sum_{n> N}x_n^2<{\varepsilon}$$
But now all is easy: consider the element $x'=(x_1,\ldots,x_N,0,0,0,\ldots)\in c_{00}$. Then $$\lVert x-x'\rVert^2 =\sum_{n> N}x_n^2<{\varepsilon}$$
Thus $c_{00}$ is dense in $\ell_2$. Since density is transitive, we conclude $\bar r=\ell_2$, as desired. $\blacktriangleleft$
Observe this is easily adapted to prove $\ell^p$ is separable for each $p\geqslant 1$, for $r\subseteq c_{00}\subseteq \ell^p$ and $r$ is countable: it is essentially the same as $$\bigcup_{k\geqslant 1} \Bbb Q^k$$
Best Answer
That is actually a bit simpler. It suffices to see that since $x \in l_p$ , $\sum |x_n|^p \leq \infty$. Therefore, this series converges, which implies that $ \forall \ \epsilon > 0$, there exists $N_0$ such that $\sum_{n\geq N} |x_n|^p < \epsilon $ for all $n \geq N_0$. Then let $S_N = (x_1,\cdots,x_N,0,\cdots)$, what we have just asserted garantees that $\|x-S_N\|<\epsilon$ $\forall \ N \geq N_0$.