The width of a convex body is a continuous function

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I'm studying an article by Shinzo Kakutani.

https://www.jstor.org/stable/1968964?origin=crossref

It states that every convex body $K$ (a compact convex set with non-empty interior) in 3-dimensional space has a circumscribed cube, i.e. a cube all of whose faces touch the body. In the proof of the theorem he defines a function $f:\mathbb{S}^2⟶\mathbb{R}$ such that for any $v\in\mathbb{S}^2$ there is a unique support hyperplane $H(K,v)$ of $K$ with outer normal vector $v.$

Let $f(v)$ be the vertical distance of these two planes $H(K,v)$ and $H(K,-v)$, $f(v)=dist(H(K,v),H(K,-v))$ I proved that $f$ is a well-defined function, guaranteeing the existence of such planes in the definition but I can't prove that $f(v)$ is a continuous function. How to calculate the $\lim_{w\to v}|f(w)−f(v)|$?

I know that $f$ is a width function, but I can't find this demonstration of continuity.

I would be very grateful for any suggestions.