$|Ax|≤\|A\||x|\space\forall x\in\mathbb{R}^n$ (Rudin's Principles) In Rudin's Principles of Mathematical analysis p. 208
$\|A\|$ is defined as the $\sup$ of all numbers $|Ax|$, where $x$ ranges over all vectors in $\mathbb{R}^n$ with $|x|≤1$.
Then he claims
$$|Ax|≤\|A\||x|$$
$\forall x\in\mathbb{R}^n$.
And uses this to prove $$\|AB\|≤\|A\|\|B\|$$
I'm confused about the mixing of $|\cdot|$ and $\|\cdot\|$ here and what then do the statements mean. What is $|\cdot|$ of a vector?
Also,
Why exactly does 
$$|Ax|≤\|A\||x|$$
hold?
 A: Recall that if $x$ is an element  of $\mathbb{R}^{n}$ then $$|x|= \sqrt{\sum_{k=1}^{n}x_k^2}$$
where $x_k$ is the $k$th component of $x$.
Recall also for any real number $\alpha$ that $|\alpha x|=|\alpha||x|$ where $|\alpha|$ is the absolute value of $\alpha$, and hence $|A(\alpha x)|=|\alpha(Ax)|=|\alpha||Ax|$
By definition $\|A\|=\sup\{|Ax|\,|\,x \in \mathbb{R}^n, |x| \leq 1\}$ and hence for any $y$ in $\mathbb{R}^n$ such that $|y| \leq 1$ we must have by the definition of supremum that $|Ay| \leq \|A\|$.
Now lets prove for all $x$ in $\mathbb{R}^n$ that $|Ax| \leq \|A\||x|$.
If $x= 0$, then this is trivially true (both sides are $0$). Now suppose $x\neq 0$ and let $y= \frac{1}{|x|}x$. It follows that $|y| = 1$ and hence $|Ay| \leq \|A\|$ (as explained above). But we also have that $|Ay|=|A(\frac{1}{|x|}x)|=\frac{|Ax|}{|x|}$. Therefore (since $|x|>0$) we have $$|Ax| = |Ay||x| \leq \|A\||x|$$ 
as required.
Let us use this to prove $\|AB\| \leq \|A\|\|B\|$.
Now suppose that $x$ is in $\mathbb{R}^n$ and $|x| \leq 1$ it follows that
$$\begin{array}{lll}|(AB)x|&=|A(Bx)| \leq \|A\||Bx| &\qquad \text{using what we proved above (with vector $Bx$)}\\  &\leq \|A\|\|B\||x| &\qquad \text{using what we proved above a second time}\\
&& \qquad \text{and multiplying by $\|A\|$}\\ &\leq \|A\|\|B\| & \qquad \text{since $|x| \leq 1$}\end{array}$$
From this we see that $\|A\|\|B\|$ is an upper bound for the set $\{|(AB)x|\,|\,x \in \mathbb{R}^{n}, |x| \leq 1\}$ and hence $$\|AB\| = \sup\{|(AB)x|\,|\,x \in \mathbb{R}^{n}, |x| \leq 1\} \leq \|A\|\|B\|$$
A: This is a general property of bounded (linear) operators on normed spaces; if $X$ is a such space and $A:X\to X$ is a bounded operator then given $y\in X$, we have
$$\|Ax\| = \left\|A\frac x{\|x\|} \right\|\|x\|\leqslant \|A\|\|x\|, $$ 
as $$\left\|\frac x{\|x\|}\right\|=1. $$
It follows that if $A$ and $B$ are bounded operators on $X$, then if $\|x\|=1$,
$$\|ABx\|\leqslant \|A\|\|Bx\|\leqslant \|A\|\|B\|\|x\| = \|A\|\|B\|, $$
so that $\|AB\|\leqslant \|A\|\|B\|$.
