Dual set of the unit ball is part of the unit ball. Define the unit ball centered at the origin as $B=\{x\in\mathbb{R}^d\mid \|x\|\leq 1\}$.
Define the dual set of set $X$ as $X^*=\{y\in\mathbb{R}^d\mid\langle x,y \rangle\leq 1\ \forall x\in X\}$.
I'm attempting to prove that for any set $C\subseteq\mathbb{R}^d$ it holds that $C=B$ if and only if $C=C^*$. I've managed to prove the implication from the right to the left by the Cauchy-Schwarz inequality.
However, I'm having some trouble with the converse implication. In particular, how to prove that $C=B$ implies $C^*\subseteq C$?

Ok, I think I got it. Thanks to @AhmedHussein for the hint.
Take any $y\in B^*$. If $y=0$, then clearly $y\in B$, so assume otherwise. Notice that $\frac{1}{||y||}y\in B$, and thus by the definition of $B^*$ it holds that $\langle\frac{1}{||y||}y, y\rangle\leq 1$.
Extract the coefficient: $\frac{1}{||y||}\langle y, y\rangle\leq 1$.
Multiply both sides by $||y||$: $\langle y, y\rangle\leq ||y||$
We can now substitute $||y||^2=\langle y,y \rangle$ and obtain inequality: $||y||^2\leq ||y||$. This is true only if $||y||\leq 1$. Hence, $y$ belongs to $B$.
