Let $(X,|\cdot|)$ be a real Banach space and $K\subseteq X$ a closed cone, that is, $K$ is a nonempty closed subset of $X$ satisfying: (a) $0\in K$; (b) $x+y\in K$ whenever $x,y\in K$; (c) $tx\in K$ whenever $t>0$ and $x\in K$; (d) if $x\in K\setminus\{0\}$ then $-x\notin K$. Denote $K_1=\{x\in K : |x|=1\}$.
Question: Is it true that there exist a closed (affine) subspace $Y\subseteq X$ and a convex set $C\subseteq Y$ such that $K_1$ is homeomorphic to $C$?
I'm reading the paper "On Krasnoselskii’s Cone Fixed Point Theorem" by Kwong [2008] and the author simply say it's easily seeing, so I guess it's just something I'm not looking at in the right way. In fact, it's pretty intuitive, and I'm looking for a formal proof. Thanks in advance.
My thoughts: It's easy to see that $K$ is necessarily convex. Then the closed convex hull of $K_1$, $\operatorname{conv}(K_1)$, is contained in $K$. I've tried to prove that $0$ does not belong to $\operatorname{conv}(K_1)$ in order to use the Hahn Banach Separation Theorem to get a functional $\phi\in X^*$ and $\alpha>0$ such that $\phi(x)>\alpha$ for every $x\in\operatorname{conv}(K_1)$. This would give me an affine subspace $A=\{x\in X : \phi(x)=\alpha\}$ which separates $K_1$ and $\{0\}$. Next, I would like to define $C=A\cap K$ and the homeomorphism I'm looking for could be a radial projection or something similar.
There are very steps that seem unclear to me in this argument, I'd appreciate any comments, hints or improvements.