[proofplan]
We prove the equality by expanding the definition of the convex subdifferential for the extended-valued indicator function. Since $x \in C$, the value of the indicator at $x$ is $0$. The subgradient inequality is automatic outside $C$ because the indicator is $+\infty$, while on $C$ it becomes exactly the defining inequality for the normal cone.
[/proofplan]
[step:Expand the subgradient inequality for the extended-valued indicator]
Fix $x \in C$ and let $v \in \mathbb{R}^n$. By definition of the convex subdifferential,
\begin{align*}
v \in \partial \mathbb{1}_C^\infty(x)
\end{align*}
if and only if, for every $y \in \mathbb{R}^n$,
\begin{align*}
\mathbb{1}_C^\infty(y) \geq \mathbb{1}_C^\infty(x) + v \cdot (y - x).
\end{align*}
Since $x \in C$, the definition of $\mathbb{1}_C^\infty$ gives $\mathbb{1}_C^\infty(x) = 0$. Hence the condition is equivalent to
\begin{align*}
\mathbb{1}_C^\infty(y) \geq v \cdot (y - x)
\end{align*}
for every $y \in \mathbb{R}^n$.
[/step]
[step:Restrict the inequality to points of $C$]
For $y \notin C$, one has $\mathbb{1}_C^\infty(y) = +\infty$, so the inequality
\begin{align*}
\mathbb{1}_C^\infty(y) \geq v \cdot (y - x)
\end{align*}
holds automatically because $v \cdot (y - x) \in \mathbb{R}$. Therefore only points $y \in C$ impose a condition. For such $y$, one has $\mathbb{1}_C^\infty(y) = 0$, and the subgradient inequality becomes
\begin{align*}
0 \geq v \cdot (y - x).
\end{align*}
Thus
\begin{align*}
v \in \partial \mathbb{1}_C^\infty(x)
\end{align*}
if and only if
\begin{align*}
v \cdot (y - x) \leq 0
\end{align*}
for every $y \in C$.
[guided]
Fix $x \in C$ and fix a vector $v \in \mathbb{R}^n$. We want to decide whether $v$ belongs to the subdifferential of the extended-valued convex function $\mathbb{1}_C^\infty$ at $x$. By the definition of the convex subdifferential, this means precisely that the affine function with slope $v$ and base value $\mathbb{1}_C^\infty(x)$ lies below $\mathbb{1}_C^\infty$ everywhere on $\mathbb{R}^n$:
\begin{align*}
\mathbb{1}_C^\infty(y) \geq \mathbb{1}_C^\infty(x) + v \cdot (y - x)
\end{align*}
for every $y \in \mathbb{R}^n$.
Because $x \in C$, the extended indicator has value $0$ at $x$. Therefore the inequality reduces to
\begin{align*}
\mathbb{1}_C^\infty(y) \geq v \cdot (y - x)
\end{align*}
for every $y \in \mathbb{R}^n$.
Now split the verification according to whether $y$ lies in $C$. If $y \notin C$, then $\mathbb{1}_C^\infty(y) = +\infty$. Since $v \cdot (y - x)$ is a finite real number, the inequality
\begin{align*}
+\infty \geq v \cdot (y - x)
\end{align*}
is automatically true and imposes no restriction on $v$.
If $y \in C$, then $\mathbb{1}_C^\infty(y) = 0$. For these points the same inequality becomes
\begin{align*}
0 \geq v \cdot (y - x).
\end{align*}
Thus the subgradient condition is equivalent to the single family of inequalities
\begin{align*}
v \cdot (y - x) \leq 0
\end{align*}
for every $y \in C$. This is exactly the condition that $v$ form a nonacute outward normal inequality against every chord $y - x$ of the convex set based at $x$.
[/guided]
[/step]
[step:Identify the resulting set with the normal cone]
By the definition of the normal cone stated in the theorem,
\begin{align*}
N_C(x) = \{v \in \mathbb{R}^n : v \cdot (y - x) \leq 0 \text{ for every } y \in C\}.
\end{align*}
The previous step showed that a vector $v \in \mathbb{R}^n$ belongs to $\partial \mathbb{1}_C^\infty(x)$ if and only if it satisfies this defining condition. Hence
\begin{align*}
\partial \mathbb{1}_C^\infty(x) = N_C(x).
\end{align*}
Since $x \in C$ was arbitrary, the equality holds for every $x \in C$.
[/step]
