Proving the definition of $\|x\|$ Proving the definition of $\|x\|$  for $x$ in a normed space $X$. I am told: $$\|x\| = \sup_{f \in X^{\star}, \|f\| = 1 }|f(x)|$$

attempt:

to prove this I know that $$\|f\| = \sup_{x \not= 0} \frac{|f(x)|}{\|x\|}$$which means for all $x \not = 0$ we have that  $\dfrac{|f(x)|}{\|x\|} \leq \|f\|$. Rearranging gives $$\|x\| \geq \dfrac{|f(x)|}{\|f\|}, \forall x \in X$$ since this holds for all x, is it enough to jump to the conclusion that $$\|x\| = \sup_{f \in X^{\star}, \|f\| = 1 }|f(x)|$$
this only poof that $\| x |\ \geq \sup_{\|f\|=1}\| f(x) \|$, for the converse define $f(\alpha x)=\alpha \| x\| \;\; for \;  x\not =0$ and use Hahn- Banach Theorem
Given
$$\|f\| = \sup_{x \not= 0} \frac{|f(x)|}{\|x\|}$$
you can multiply by $\|\tilde x\|$ on both sides and get
$$\|\tilde x\|\|f\| = \|\tilde x\|\sup_{x \not= 0} \frac{|f(x)|}{\|x\|}$$
This equation holds for all $\tilde x$, so it also holds for the one $x$ in the supremum term. So you can rewrite this equation to
$$\|x\|\|f\| = \sup_{x \not= 0}\|x\|\frac{|f(x)|}{\|x\|}$$
which clearly is
$$\|x\|\|f\| = \sup_{x \not= 0}|f(x)|$$
By setting $\|f\|$ to $1$ you are finished.