Is the inclusion map of $\ell^1(N)$ in $\ell^2(N)$ bounded and dense? I am looking for an idea to prove if the inclusion map from $\ell^1(N)$ to $\ell^2(N)$ bounded and does it have a dense image.
And why is the set $A:=\{x: ||x||_1\le 1\} \subset \ell^2(N)$ closed and nowhere dense with the norm topology on $\ell^2(N)$?
 A: As for the proof of continuity of natural inclusion $i:x\mapsto x$ of $\ell_1(\mathbb{N})$ into $\ell_2(\mathbb{N})$ see this answer. In your case $p=1$, $q=2$. Note that $i$ maps finitely supported sequences of $\ell_1(\mathbb{N})$ onto finitely supported sequences of $\ell_2(\mathbb{N})$ so they are in the image . Recall this sequences are dense in $\ell_2(\mathbb{N})$. Hence $\mathrm{Im}(i)$ is dense in $\ell_2(\mathbb{N})$. As for the second question take a look at this answer.
A: Let $x\in\ell^1$, then $|x(n)|<1$ for $n\geqslant n_0$, hence $|x(n)|^2<|x(n)|$ for these $n$ and this proves that $\ell^1\subset\ell^2$. We have that 
$$\lVert x\rVert_2^2=\sum_{j=0}^{+\infty}|x(n)|^2\leqslant\left(\sum_{j=0}^{+\infty}|x(j)|\right)^2=\sum_{j=0}^{+\infty}|x(n)|^2+2\sum_{l\leq k}|x(l)|\cdot |x(k)|,$$
which proves the boundedness of the inclusion. 
As the sequences of finite support are dense in $\ell^2$, the image is dense. 
$A$ is closed (in $\ell^2$). Indeed, if $x^n\to x$ in $\ell^2$ and $\lVert x^n\rVert_1\leqslant 1$ for all $n$, then for each $n$ and each integer $N$, 
$$\sum_{j=0}^N|x_j|\leqslant \sum_{j=0}^N|x_j^n|+\sum_{j=0}^N|x_j-x_j^n|\leqslant 1+\sum_{j=0}^N|x_j-x_j^n|\leqslant 1+\sqrt N\lVert x-x^n\rVert_2.$$
Take the $\lim_{n\to+\infty}$ on both sides to get $\sum_{j=0}^N|x_j|\leqslant 1$ for all $N$, giving the wanted result. 
The fact that $A$ has non-empty interior is done here.
