Prove that there is no fixed point Let $f: \mathbb R \to \mathbb R$, whereby $f(x):=(x+\sqrt{x^{2}+1})/2$ with $x \in \mathbb R$. 
Show that $f$ fulfils: $\forall x, y \in \mathbb R$ with $x \neq y$ 
$d(f(x),f(y))<d(x,y)$, whereby $d(x,y):=|x-y|$
(First question they have not mentioned the metric $d$ used, so how can I prove this? Must I just conclude that we are using the euclidean norm?)
Then show that $f$ does not have a single fixed point. (I'm assuming it's easiest to prove this via contradiction). 

I'm assuming that $d$ is the euclidean metric.
We have $f'(x) = \frac12\left(1 + \frac{x}{\sqrt{x^2+1}}\right)$ so by the mean value theorem there exists $c$ between $x$ and $y$ such that
$$|f(x) - f(y)| \le |f'(c)||x-y| < |x-y|$$
because for all $c \in \mathbb{R}$ holds
$$|f'(c)| \le \frac12\left(1 + \frac{|x|}{\sqrt{x^2+1}}\right) < \frac12(1+1) = 1$$
However, $f$ does not have a fixed point:
$$x = f(x) = \frac12\left(x + \sqrt{x^2 + 1}\right) \implies x = \sqrt{x^2 + 1}$$
which is impossible because $|x| < \sqrt{x^2 + 1}$ for all $x \in \mathbb{R}$.
