$\newcommand{\R}{{\mathbb{R}}}\newcommand{\Rbar}{\overline{\R}}\newcommand{\<}{\langle}\newcommand{\>}{\rangle}\newcommand{\dom}{\operatorname{dom}}$
Let me give some definitions for completeness. The subdifferential of a function $f:\R^n\to\Rbar$ at $x$ is defined as
$$
(\partial f)(x) := \{v \in \R^n {}:{} f(x') \geq f(x) + \<v, x'-x\>, \forall x' \in \R^n\}.
$$
The recession cone of a set $X \subseteq \R^n$ - allow me to denote it by $C_{\infty}$ - is the set
$$
X_\infty = \{d \in \R^n {}:{} x+ad \in X, \forall x\in\R^n, \forall \lambda\geq 0\}.
$$
For a function $f:\R^n\to\Rbar$, its horizon function is defined as
$$
f_{\infty}(x) = \liminf_{x'\to x, t \to \infty}\frac{f(tx')}{t}.
$$
The domain of $f$ is defined as $\dom f = \{x\in \R^n{}:{} f(x) < \infty\}$.
Lastly, the normal cone of a convex set $C\subseteq \R^n$ at a point $x\in\R^n$ is defined as $N_C(x) = \{y \in \R^n {}:{} \<y, x'-x\>\leq 0, \forall x'\in C\}$.
The subdifferential of $f$ at $x$ can be written as
\begin{align}
(\partial f)(x) {}={}& \bigcap_{x'\in\dom f}
\{v \in \R^n {}:{} f(x') \geq f(x) + \<v, x'-x\>\}
\\
{}={}&
\bigcap_{x\in\dom f}
\{v \in \R^n {}:{} f(x) - f(x') + \<v, x'-x\> \leq 0\}
\end{align}
Define $\phi_{x'}(v;x) = f(x) - f(x') + \<v, x'-x\>$ and $C_{x'} = \{v \in \R^n {}:{} \phi_{x'}(v;x) \leq 0\}$, therefore, the recession cone of $(\partial f)(x)$ is
\begin{align}
(\partial f)(x)
{}={}
\bigcap_{x\in\dom f} C_{x'},
\end{align}
therefore,
\begin{align}
((\partial f)(x))_{\infty}
{}={}
\left(\bigcap_{x'\in\dom f} C_{x'}\right)_{\infty}
{}={}\bigcap_{x'\in\dom f} (C_{x'})_{\infty}
\end{align}
We will not use the fact that $(C_{x'})_{\infty} = \{y\in \R^n {}:{} (\phi_{x'})_{\infty}(v;x) \leq 0, \forall x' \in \dom f\}$, but
$$
(\phi_{x'})_{\infty}(v;x) = \lim_{x''\to x', t\to\infty}
\frac{f(x) - f(x'') + \<tv, x''-x\>}{t}
=
\<v, x'-x\>,
$$
so $C_{x'} = \{v: \<v, x'-x\> \leq 0\}$ [this is Exercise 3.24 in the book of Rockafellar and Wets, Variational Analysis, Springer]. Then,
\begin{align}
((\partial f)(x))_{\infty}
{}={}\bigcap_{x'\in\dom f} \{v: \<v, x'-x\> \leq 0\}
{}={}N_{\dom f}(x)
{}={}N_{\mathbb{C}}(x).
\end{align}
Since $x$ is at the boundary of $C$, it is a cone which is a proper subset of $\R^n$.
