Why is $||x-y||=\sup_{|| f|| \leq 1} |f(x-y)|$? Why is $||x-y||=\sup_{|| f|| \leq 1} |f(x-y)|$?
Where $x,y$ are in normed space $V$ and $f$ is a linear bounded functional in $V'$, the dual of $V$.
As given here p.5:
http://courses.mai.liu.se/GU/TATM85/handout-15.pdf

Perhaps they consider the dual norm definition in "reverse" going from $X'$ to $X$, rather than $X$ to $X'$?
https://en.wikipedia.org/wiki/Dual_norm#Definition
But this would require commutitativity?
 A: Note that for every bounded linear functional $f$ we have 
$$|f(x)-f(y)|=|f(x-y)|\leqslant ||f||_{V\to V'} ||x-y||_V$$
where $||\cdot||_{V\to V'}$ is the operator norm and $||\cdot||_V$ is the norm in $V$. Therefore
$$\frac{|f(x-y)|}{||f||_{V\to V'}}\leqslant ||x-y||_V\Rightarrow\sup_{||f||\leqslant 1}\frac{|f(x-y)|}{||f||_{V\to V'}}=\sup_{||f||\leqslant 1}|f(x-y)|\leqslant \sup_{||f||\leqslant 1}||x-y||_V=||x-y||_V$$
So we get one side of the inequality
$$||x-y||_V\geqslant \sup_{||f||\leqslant 1}|f(x-y)|$$
We also need to show that $$\sup_{||f||\leqslant 1}|f(x-y)|\geqslant ||x-y||_V$$
It suffices to have a functional $f$ such that $||f||_{V\to V'}=1$ and such that $|f(x-y)|=||x-y||_V$. But existence of such a functional follows from a fundamental fact in functional analysis (as a corollary of Hahn-Banach Theorem).
$\textbf{Corollary:}$ For every $x\in V\setminus\{0\}$ there exists a bounded linear functional $f:V\to V'$ such that $||f||_{V\to V'}=1$ and $f(x)=||x||_V$. 
$\textbf{Theorem:}$ Let $V$ be a normed space and $U\subseteq V$ a subset. Let $x\in V$ satisfy $$d:=\inf_{y\in U}||y-x||_V>0$$
Then there exists some bounded linear functional $f:V\to V'$ such that
$f(y)=0$ for all $y\in U$, $f(x)=d$ and $||f||_{V\to V'}=1$.
The proof of this last theorem involves Hahn-Banach theorem (the technique of extending a linear functional) which I am not going to write here (you can find it in any standard text of functional analysis). Note that the corollary follows immediately from this theorem with $U:=\{0\}$.
