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?

By Cauchy-Schartz inequality for $v,w \in V$ you have $\vert \langle v,w \rangle \vert \le \Vert v \Vert \Vert w \Vert$. Hence for $\Vert w \Vert =1$ you have $\vert \langle v,w \rangle \vert \le \Vert v \Vert$. Which proves that $\sup\{\vert \langle v,w \rangle \vert | \ w \in V, \ \Vert w \Vert =1\} \le \Vert w \Vert$
To prove the reverse inequality you know that $\Vert v \Vert^2=\langle v,v \rangle = \Vert v \Vert \langle v,\frac{v}{\Vert v \Vert} \rangle$ where $\frac{v}{\Vert v \Vert}$ is a vector with norm equal to $1$ in the case $v \neq 0$.
And the equality is trivial when $v = 0$.
