The problem is to show $\ell^2$ has a countable dense subset. We have proved in an earlier problem that the set $U = \{x \in \mathscr \ell ^2\ \mid \exists n \in \mathbb N : \forall n \ge N : x_n = 0\}$ (the first $N-1$ coordinates of $x$, the rest are zero) is dense in $\ell^2$, that is for fixed $x \in \ell^2$ there exists $u \in U$ arbitrarily close to $x$.
I think if we define $Y_i$ to be the (countable) set of successive rational approximations to $u_i$ (since $\mathbb Q$ is dense in $\mathbb R$), then the finite Cartesian product $Y_1 \times \cdots \times Y_{N-1}$ will have some $y$ with coordinates arbitrarily close to the coordinates of $u$ which is arbitrarily close to $x$, and also be countable. (The notation is wrong since the Cartesian product actually represents a sequence in $\ell^2$ with finitely many terms followed by infinite zeros)
Is this the right idea? How can I formalize this idea more?
Consider the set $Y_0 \subset \ell^2$, it consists of sequences $(y_1, y_2, \dots, y_n, 0, \dots)$, where $y_k \in \mathbb{Q}$. This set is countable, because it is the union of a countable number of countable sets $\mathbb{Q^n}$. Now consider the sequence $l=(l_1, l_2, \dots ,l_n, l_{n+1}, \dots) \in \ell^2$. Obviously, $\exists \varepsilon > 0$: $$\sum_{k=n+1}^{\infty}l_k^2 < \frac{\varepsilon^2}{2}$$
Since the set $\mathbb{Q}$ is everywhere dense in $\mathbb{R}$, then in the interval $$\Big(l_i - \frac{\varepsilon}{\sqrt{2N}}, l_i + \frac{\varepsilon}{\sqrt{2N}}\Big)$$ exist a rational number $q_i \in \mathbb{Q}$. Obvious, that $$|l_i - q_i|^{2} < \frac{\varepsilon^2}{2N}$$ Consider a sequence: $q = (q_1, q_2, \dots, q_n, 0, \dots) \in Y_0$ $$\rho(l, q) = \sqrt{\sum\limits_{k = 1}^{\infty} |l_{k} - q_{k}|^{2}} = \sqrt{ \sum_{k = 1}^{n} |l_{k} - q_{k}|^{2} + \sum_{k = n+1}^{\infty} |l_{k}|^{2}} < \sqrt{\frac {n \cdot \varepsilon^{2}}{2n} + \frac {\varepsilon^{2}}{2}} =\varepsilon.$$
We show that $q \in B(l, \varepsilon) \Rightarrow Y_0$ is everywhere dense in $\ell^2$.
Similarly, you can prove the separability $\ell^p$.