Take elements and notation as in your post, then if \(\displaystyle (x_n) \in L ^p\) then, eventually, \(\displaystyle |x_n|<1\), and note that the function \(\displaystyle f:[1,\infty ) \rightarrow \mathbb{R}\) given by \(\displaystyle f(t)=a^t\) is decreasing iff \(\displaystyle a\in (0,1)\) and so, eventually, for \(\displaystyle n\) big enough you get \(\displaystyle |x_n|^s \leq |x_n|^p\).

Another interesting question would be to see if the inclusion mapping is bounded.

That's exactly why I can't conclude the boundedness of the inclusion mapping, but as far as set inclusion goes it suffices because only finitely many terms aren't dominated.

I think I have an answer so if anyone's interested here it goes:

Since the inclusion is clearly linear we only have to prove that it's continous at \(\displaystyle 0\). Take \(\displaystyle (x_n)_m \subset L^p\) such that \(\displaystyle (x_n)_m \rightarrow 0\) as \(\displaystyle m\rightarrow \infty\). There exists an \(\displaystyle M\) such that if \(\displaystyle m>M\) then \(\displaystyle \| (x_n)_m \|_p ^p <1\) so we must have \(\displaystyle |x_{n,m} | <1\) for all \(\displaystyle n\in \mathbb{N}\) and \(\displaystyle m>M\), but by the argument used to prove the inclusion we then have \(\displaystyle 0\leq \| (x_n)_m\| _s^s\leq \| x\| _p^p \rightarrow 0\) so that it is continous at \(\displaystyle 0\).