Prove that sequence space $\ell_p(\mathbb R)$ is separable

To elaborate on above answer, let $M$ be the set of all sequences $(r_{0}, \ldots, r_{n}, 0,\ldots)$ with $r_{i}$ being rational and $n \in \mathbb{N}$. (This is the set of sequences with finite support and rational entries).

Since $\mathbb{Q}$ is countable and finite product of countable sets is countable and then countable union of countable sets is countable, we see that $M$ is countable.

Now let us see that $M$ is dense in $\ell_{p}(\mathbb{R})$. To do this, given any $x = (x_{n}) \in \ell_{p}(\mathbb{R})$, and for $\epsilon > 0$ we must find an element $y \in M$ such that $d(x,y) <\epsilon$. We have

$$ \sum\limits_{n=0}^{\infty} |x_{n}|^{p} < \infty $$

Hence, given $\epsilon > 0$, there exists $m \in \mathbb{N}$ such that

$$ \sum\limits_{n=m+1}^{\infty} |x_{n}|^{p} < \epsilon/2 $$

Now for $0\leq i \leq m$, choose $r_{i} \in \mathbb{Q}$ such that $|r_{i} - x_{i}| < \left(\frac{\epsilon}{2m}\right)^{\frac{1}{p}}$ (using that the rationals are dense in reals).

Then the element $y = (r_{0}, r_{1}, \ldots, r_{m}, 0, 0, \ldots) \in M$ is the required element.

The sequences with finite support and rational entries are dense in $\ell_p(\mathbb R)$ for each $1\leqslant p\lt \infty$ (we can make the remainder of the series as small as we wish).