Let $C = \text{cone}(u_1,\dots,u_m)$ for some $u_1,\dots,u_m \in \mathbb{R}^d \setminus \{\textbf{0}\}$ be a finitely generated pointed cone. Let $H_0 := \{x \in \mathbb{R}^d: \langle a,x \rangle = 0\}$ be a hyperplane such that $C \subseteq H_0^\geq$ and $C \cap H_0 = \{\textbf{0}\}$. For some $\delta > 0$, let $H_\delta := \{x \in \mathbb{R}^d: \langle a,x \rangle = \delta \}$ and $C_\infty := C \cap H_\delta$.
I want to show that $C_\infty$ is a polytope. Since both $C$ and $H_\delta$ could be represented by a system of inequalities, so is $C_\infty$, which shows that it is a polyhedron. Now I just need to show that it is also bounded.
I was attempting a proof by contradiction, assuming that $C_\infty$ is not bounded, hence containing a ray $\{x + ty: t \geq 0\}$ for some $x \in C_\infty$ and $y \neq \textbf{0}$ (and in particular a point $q = x + ty, t > 0$ on this ray). This should lead to $C \cap H_0$ contains more than just $\textbf{0}$, but I haven’t been able to show it so far. I know that $y$ as defined above should be in $H_0$, but that has not helped.
How about this.
If $x\in H_\delta\cap C$, then $\epsilon x\in H_{\epsilon\delta}\cap C$ for every $\epsilon>0$.
That is, if for some $y\in H_0\setminus\{0\}$ and all $t\ge 0$ holds $x+ty\in H_\delta\cap C$, then also $\epsilon x + ty\in H_{\epsilon\delta}\cap C$ for all $t\ge 0$ (the $\epsilon$ in the last term got absorbed into the $t$). Note that $C$ can be written as the union
$$C=\bigcup_{\epsilon \ge 0} (H_{\epsilon\delta}\cap C),$$
and so $\epsilon x + t y\in C$ for all $\epsilon > 0, t \ge 0$. But since $C$ is a closed set, with $\epsilon\to 0$ we find $yt\in C$ for all $t\ge 0$. In particular, for $t>0$ and since $y\in H_0\setminus\{0\}$, we found $0\not= yt\in H_0\cap C$.
The crucial part is indeed that $C$ is closd, because the statement is wrong for open cones.