= Solution
A subset is <nowhere dense> when its closure has empty interior. It is of first category, or meagre, when it is a <countable union> of nowhere-dense subsets; otherwise it is of second category.
The <Baire category theorem> says that in a <complete metric space> every <countable intersection> of open <dense subsets> is dense. To prove it, start inside an arbitrary nonempty <open set> and choose a <closed ball> of positive radius there and in the first dense <open set>. Inductively choose a <closed ball> inside the interior of the preceding ball and the next dense <open set>, with radius less than $2^{-n}$. The centres are Cauchy. Completeness gives a limit lying in all the nested <closed balls>, hence in all the dense <open sets> and the original <open set>. Thus a nonempty <complete metric space> is not meagre in itself.
For $p<r\le\infty$, put $E_m=\{x\in\ell^p:\|x\|_p\le m\}$, viewed inside $\ell^r$. Necessarily $p<\infty$. Each $E_m$ is closed: convergence in the $\ell^r$ <norm> gives coordinatewise convergence, and every finite <partial sum> of $\sum|x_i|^p$ is bounded by $m^p$ in the limit. Taking the supremum of <partial sums> proves the assertion.
It has empty interior. Given a point in it and any small $\ell^r$ radius, perturb a block of $N$ coordinates by entries of modulus $\delta N^{-1/r}$, choosing signs to avoid cancelling the existing coordinates. The $\ell^r$ distance is $\delta$, while the resulting $\ell^p$ <norm> is at least $\delta N^{1/p-1/r}$, exceeding $m$ for large $N$. For $r=\infty$, use entries of modulus $\delta$ on that block. Hence each $E_m$ is <nowhere dense>, and
$$
\boxed{\ell^p=\bigcup_{m=1}^\infty E_m\text{ is first category in }\ell^r\ (p<r).}
$$
For $r=p$, including $p=\infty$, the space is Banach and so \b[second category in itself] by Baire. There is no case $r>p$ when $p=\infty$.
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