If $x=(x_n) \in \ell^p$ with $\|x\|_p = 1$, then $|x_n| \le 1$ for all $n$, so $|x_n|^p \le |x_n|^q$ for all $n$, and so $\|x\|_q \le 1$. Using $x=(1,0,0,\ldots)$, it is easy to see that the operator norm of the embedding of $\ell_p$ into $\ell_q$ is actually equal to $1$.
As to the second question, the sequence $(1,1,1,\ldots) \in \ell^\infty$ is not in the direct limit of the $\ell^p$-spaces, which suggests that the answer is no. In order to show that they are not homeomorphic, you can use the fact that $\ell^\infty$ is not separable, but each $\ell^p$ for $1\le p < \infty$ is, so the direct limit is separable, too. (Just take the image of countable dense sets in $\ell^1, \ell^2, \ell^3, \ldots$ in the direct limit, giving you a countable dense set in the direct limit.)