I know the definitions:
We say the sequence $(a_{n})$ belongs to $l^p$ if $$\left(\sum_{n=1}^{\infty}|a_n|^p\right)^{\frac 1p}<\infty. $$
So, the unit ball of $l^1$, and I will denote this by $B_{1},$ is the set of all sequences $(a_n)$ satisfying $$ \sum_n |a_ n|\leq1.$$
To prove that $B_1$ has empty interior I think I need use the Baire's theorem, for $l^p$ is complete. So, how can I write $B_1=\bigcup_{n\in\mathbb{N}}A_n$, with $int\,A_n=\emptyset$?
I will answer the question in the title which says that the unit ball of $\ell ^{1}$ has no interior in $\ell ^{2}$. Suppose $(a_n)$ is an interior point. Then there exists $\epsilon >0$ such that $\sum |b_n-a_n|^{2} <\epsilon$ implies $\sum |b_n| \leq 1$. Choose $N$ such that $\sum_{n=N}^{\infty} \frac 1 {n^{2}} <\epsilon$ an d $\sum_{n=1}^{N-1} \frac 1 n >1+\sum |a_n|$. Let $b_n=a_n+\frac 1 n$ for $n \geq N$ and $b_n=a_n$ for $n <N$. Note that $\sum |b_n-a_n|^{2} <\epsilon$ but $\sum |b_n| \geq \sum_{n=1}^{N-1} \frac 1 n -\sum_{n=1}^{N-1} |a_n|>1 $. So we have a contradiction.