How is the subdifferential of the $l_2$ norm at $x=0$ the polar of the unit ball?

I was reviewing subdifferentials, and my professor writes on the board that the subdifferential of the $l_2$ norm at $x=0$ is the polar of the unit ball?

I tried to work through this myself:

Let $\Phi(x):\mathbb{R}^n\rightarrow\mathbb{R}$ and $C$ is a closed, non-empty convex set. $$\Phi(x)=||x||$$

I know the following: $$\Phi^*(z)=\sup_{x\in C}\{\langle z,x\rangle-\Phi(x)\}$$

and I know that the support function of the set $C$ is the dual of the indicator function:

$$\sigma_C(x)=\delta^*_C(x):= \left\{ \begin{array}{ll} 0 & x\in C \\ \infty &\text{else} \\ \end{array} \right. $$

and I also know that it is equal to the sup:

$$\sigma_C(x) = \sup_z\{\langle z,x\rangle\}$$

And I know the definition of the subdifferential at the point $x$ for the set $C$ is this:

$$\partial\Phi(x)=\{z\hspace{0.2cm}|\hspace{0.2cm} \Phi(y)\ge \Phi(x)+ \langle z,y-x\rangle \hspace{0.2cm}\forall y \}$$

But I do not understand how to put these definitions together to get this:

$$\partial\Phi(0)=\mathbb{B}^o$$ which is the polar of the unit ball. How are the steps used to get this?

edit :: I am also interested in how these things relate to each other: the $\sigma_C(x)$, and $\partial\Phi(x)$, (if at all) ?

By definition of the subdifferential we get for $\Phi(0)$ the following $$\partial\Phi(0)=\{z\in\mathbb{R}^n|\Phi(y)\geqslant \Phi(0)+\langle z,y-0\rangle,\forall y\in \mathbb{R}^n\}=\{z\in\mathbb{R}^n|\hspace{0.2cm}||y||\geqslant \langle z,y\rangle,\forall y\in \mathbb{R}^n\}\\=\{z\in\mathbb{R}^n|\hspace{0.2cm}||y||\geqslant \langle z,y\rangle,\forall y\in \mathbb{R}^n, ||y||<1\}\cup \{z\in\mathbb{R}^n|\hspace{0.2cm}||y||\geqslant \langle z,y\rangle,\forall y\in \mathbb{R}^n, ||y||\geqslant 1\} \\=\{z\in\mathbb{R}^n|\hspace{0.2cm}||y||\geqslant \langle z,y\rangle,\forall y\in \mathbb{R}^n, ||y||<1\}\cup\{z\in\mathbb{R}^n|\hspace{0.2cm}1\geqslant \langle z,\frac{y}{||y||}\rangle,\forall y\in \mathbb{R}^n\}=\{z\in\mathbb{R}^n|\hspace{0.2cm}1\geqslant \langle z,y\rangle,\forall y\in \mathbb{B}\}:=\mathbb{B}^o$$ Note that $\forall y\in \mathbb{R}^n\setminus\{0\}$ we have $y/||y||\in \mathbb{B}$ where $\mathbb{B}$ is the unit ball.