How do you show monotonicity of the $\ell^p$ norms?

You are right @user1736.

If $0<a\leq1$ then $$\left(\sum |a_n|\right)^a\leq \sum|a_n|^a.\tag{1}$$

Hence for $p\leq q$ we have $p/q\leq1$, and $$\left(\sum_n |x_n|^q\right)^{1/q}=\left(\sum_n |x_n|^q\right)^{p/qp}\leq \left(\sum_n |x_n|^{q(p/q)}\right)^{1/p}=\left(\sum|x_n|^p\right)^{1/p}$$

Edit: How do we prove (1) (for $0<a<1$)?

Step 1. It is sufficient to prove this for finite sequences because then we may take limits.

Step 2. To prove the statement for finite sequences it is sufficient to prove $$(x+y)^a\leq x^a+ y^a,\quad\text{ for $x,y>0$}\tag{2}$$ because the finite case is just iterations of (2).

Step 3. To prove (2) it suffice to prove $$(1+t)^a\leq 1+t^a,\quad\text{ where $0<t<1$} \tag{3}$$

Now, the derivative of the function $f(t)=1+t^a-(1+t)^a$ is given by $f'(t) = a(t^{a-1} - (1+t)^{a-1})$ and it is positive since $a>0$ and $t\mapsto t^b$ is decreasing for negative $b$. Hence, $f(t)\geq f(0)=0$ for $0<t<1$, which proves (3).

I don't think you need to prove the inequality you have in the question; that's a bit too strong. Note that $\{x_n\}\in\ell_p$ if and only if $\left(\sum|x_n|^p\right)^{1/p}$ is finite, if and only if $\sum|x_n|^p\lt\infty$. So you really just need to show that if $\sum|x_n|^p$ is finite, then $\sum|x_n|^q$ is finite, assuming $p\leq q$.

You want to remember is two things:

1. if $p\leq q$, then for $|x|>1$ you have $|x|^p\leq|x|^q$, but if $|x|<1$, then $|x|^p \geq |x|^q$.
2. $\sum_{n=1}^{\infty}a_n$ converges if and only if for every $m\geq 1$, $\sum_{n=m}^{\infty}a_n$ converges.