Norm is sup of inner products (proof). Let $V$ be a vector space with an inner product $\langle.,. \rangle$ and associated norm $|| . ||$ Then:

Could I have a proof of this fact?

Since $\Vert \cdot \Vert$ is the induced norm, we have $\Vert v \vert = \sqrt{\langle v,v \rangle}$.
If $v = 0$, then the inequality $$\Vert v \Vert \leq \sup \left\{ \langle v,w \rangle \, : \, w \in V, \Vert w \Vert = 1 \right\} $$
is trivial, so assume $0 \neq v \in V$. Note that for $w := \frac{v}{\Vert v \Vert}$ we have $\Vert w \Vert = 1$. So we calculate $$\begin{align*}\Vert v \Vert^2 &= \langle v, v \rangle = \Vert v \Vert \left\langle v, \frac{v}{\Vert v \Vert } \right\rangle = \Vert v \Vert \langle v, w \rangle \\ &\leq \Vert v \Vert \cdot \sup \left\{ \langle v,w \rangle \, : \, w \in V, \Vert w \Vert = 1 \right\} \; ,\end{align*}$$
which gives us 
$$\Vert v \Vert \leq \sup \left\{ \langle v,w \rangle \, : \, w \in V, \Vert w \Vert = 1 \right\} \; .$$
For the other direction, we know by the Cauchy-Schwarz inequality, that 
$$ \vert \langle v,w \rangle \vert \leq \Vert v \Vert \Vert w \Vert \quad \text{for all } v,w \in V \; ,$$
so we deduce that 
$$\langle v,w \rangle \leq \Vert v \Vert  \quad \text{for all $v,w \in V$ with $\Vert w \Vert = 1$} \; ,$$
and it follows that 
$$\sup \left\{ \langle v,w \rangle \, : \, w \in V, \Vert w \Vert = 1 \right\} \leq \Vert v \Vert \; .$$
This shows the equality $$\Vert v \Vert = \sup \left\{ \langle v,w \rangle \, : \, w \in V, \Vert w \Vert = 1 \right\} \quad \text{for each $v \in V$} \; .$$
