Monotonicity of $\ell_p$ norm

This answers your first question

As for the second question. Consider Holder inequality $$ \sum\limits_{i=1}^n |a_ib_i|\leq \left(\sum\limits_{i=1}^n |a_i|^{s/(s-1)}\right)^{1-1/s}\left(\sum\limits_{i=1}^n |b_i|^s\right)^{1/s} $$ with $a_i=1$, $b_i=|x_i|^r$, $s=p/r$.

For (i): Let $p>r>0$ and define $\|x\|_p = (\sum_{i} |x_i|^p)^{1/p}$.

First suppose that $\|x\|_p=1$. Then $\|x\|_p^p=1$ and $|x_i|\le 1$ for all $i$. Since $p>r$, we obtain $|x_i|^p \le |x_i|^r$ for all $i$ and, therefore,

$$1=\|x\|_p^p = \sum_{i} |x_i|^p \le \sum_{i} |x_i|^r = \|x\|_r^r.$$

Hence, $\|x\|_r \ge 1 = \|x\|_p$.

In the general case, set $y_i=x_i/\|x\|_p$, $y$ the vector with components $y_i$. Then $\|y\|_p=1$ and we can use the first part of the proof to obtain

$$ 1=\|y\|_p \le \|y\|_r= \frac{\|x\|_r}{\|x\|_p}. $$

This works for all $p>r>0$ when we define $\|x\|_p = (\sum_{i} |x_i|^p)^{1/p}$. However, for $p<1$, this does not define a norm since it is not sub-additive. See the lines after the formula you quote from Wikipedia. Sometimes the formula $\|x\|_p=\sum_{i} |x_i|^p$ is used in the case $p<1$.