A convex closed set with disconnected boundary: is it necessarily a strip? Consider the following

Problem: Suppose $X \subset \Bbb R^n$ is a closed convex set. Can we deduce that its boundary $\partial X$ is connected?

At first sight, I was thinking about spheres, so that this seemed to be true. Then I found a counterexample: the strip between two parallel hyperplanes. For example
$$X= \{ (x_1, \dots , x_n) : 0 \le x_1 \le 1\}$$
is a counterexample.
However it seems that no other counterexample can be produced. Indeed a convex set is an intersection of some half-spaces, determined by hyoerplanes, and if I pick any two non parallel hyperplanes they would intersect "forming a connection in the boundary".
Can we write down a formal proof about this?

True Question: Suppose $X \subset \Bbb R^n$ is a closed convex set whose boundary $\partial X$ is not connected. Prove that $X$ has the form $$X= \{ \mathrm x \in \Bbb R^n : a \le \mathrm v \cdot \mathrm x \le b \}$$ for some $\mathrm v \in \Bbb R^n$ and $a,b \in \Bbb R$.

 A: Yes, if $X \subset \mathbb{R}^n$ is a closed convex set, then $\partial X$ is connected or $X$ is the space between two parallel hyperplanes. (Unless you consider $\varnothing$ to be disconnected, then $X = \varnothing$ and $X = \mathbb{R}^n$ are further exceptions.)
This is rather trivial for $n = 1$, hence I assume $n > 1$. Some preliminary observations:

*

*If $\overset{\circ}{X} = \varnothing$, then $\partial X = X$ is connected.

*If $P$ is a plane in $\mathbb{R}^n$ passing through an interior point of $X$, then $\partial_P(X\cap P) = (\partial X) \cap P$.

*If $C \subset \mathbb{C}$ is a convex neighbourhood of $0$, then the map $s \colon z \mapsto \sup \{ t > 0 : tz\in C\}$ is a continuous map from $S^1$ to $(0, +\infty]$.

The first point is obvious, since $X$ is convex, hence connected. For the second point, assume for convenience that the interior point is the origin. Then for a point $b \in \partial X$ we have $tb \notin X$ for all $t > 1$ and $tb \in X$ for $t \in [0,1]$. Thus if also $b \in P$ it follows that $b$ is an adherent point of $X\cap P$ (take $t \in [0,1]$) as well as of $P \setminus X$ (take $t > 1$), hence $b \in \partial_P(X\cap P)$. The same argument works if we look at $b' \in \partial_P(X\cap P)$, $tb \in X\cap P$ for $t \in [0,1]$ and $tb \in P\setminus X$ for $t > 1$. For the last point, if $s(z) < +\infty$, consider the straight lines through $+i\varepsilon z$ and $s(z)\cdot z$ and through $-i\varepsilon z$ and $s(z)\cdot z$ for small enough $\varepsilon > 0$. For $w \in S^1$ close enough to $z$, $s(w)$ must be finite and $s(w)\cdot w$ must lie between these two straight lines, hence $s$ is continuous at $z$. If $s(z) = +\infty$, considering the triangle with vertices $\pm i\varepsilon z$ and $R\cdot z$ shows $s(w) > M$ for $\lvert w - z\rvert < \varepsilon/(3M)$ when $R$ is chosen large enough. (Alternative shorter proof of the third point: The Minkowski functional $\mu_C$ of $C$ is a continuous sub-norm on $\mathbb{C}$, and $s(z) = \frac{1}{\mu_C(z)}$.)
Now consider a closed convex $X \subset \mathbb{R}^n$ with disconnected boundary. Let $C_1, C_2$ be two connected components of $\partial X$ (we don't know yet that there are only two), and $q_i \in C_i$. The line segment connecting $q_1$ and $q_2$ cannot be contained in $\partial X$, since $q_1$ and $q_2$ lie in different components of $\partial X$. Hence with exception of the endpoints it lies entirely in $\overset{\circ}{X}$. The function given by
$$f(x) = \operatorname{dist}(x,C_1) - \operatorname{dist}(x,C_2)$$
(we use the Euclidean norm) is continuous, with $f(q_1) < 0$ and $f(q_2) > 0$ (since $C_i$ is closed in $\mathbb{R}^n$), hence there is an $x_0$ on the line segment between $q_1$ and $q_2$ with $f(x_0) = 0$. Without loss of generality $x_0 = 0$, and $\operatorname{dist}(0, C_i) = 1$. Let $p_i \in C_i$ with $\lVert p_i\rVert = 1$. Without loss of generality let $p_1 = (1,0, \dotsc, 0)$.
Then $\{x : x_1 = 1\}$ is a supporting hyperplane for $\partial X$, i.e. $x \in X \implies x_1 \leqslant 1$. For if there were $x \in X$ with $x_1 > 1$, consider the plane $P$ through $0, p_1$ and $x$. The straight line through $x$ and $p_1$ cuts off a segment from the unit disc, and that segment cannot contain points of $X\cap P$, for otherwise $p_1 \notin \partial_P(X\cap P)$. Identifying the plane with $\mathbb{C}$ we see that $s(z) < 1$ on the open arc bounding the segment, and $s(z)\cdot z$ is a path in $C_1$ containing points closer to $0$ than $p_1$. This contradiction proves the assertion that $x_1 \leqslant 1$ for all $x \in X$.
Now it follows further that $p_2 = -p_1$. For otherwise consider the plane $P$ through $0, p_1, p_2$. We would have $s(z) < +\infty$ for all $z$ on the shorter of the two arcs between $p_1$ and $p_2$, and $s(z)\cdot z$ would give a path in $\partial X$ connecting $p_1$ and $p_2$, contradicting the premise that these points lie in different components of $\partial X$.
As above we see that $\{x : x_1 = -1\}$ is a supporting hyperplane, hence $X \subset \{ x : \lvert x_1\rvert = 1\}$.
We actually have equality, i.e. $X = \{x : \lvert x_1\rvert \leqslant 1\}$.
Suppose there were a point $q_0 \in \{x : \lvert x_1\rvert \leqslant 1\} \setminus X$. Multiplying $q_0$ with a suitable $t \in (0,1)$ we obtain $q \in \partial X$ with $\lvert q_1\rvert < 1$. Now consider the plane $P$ through $0, p_1, q$. On the side of the straight line through $p_1$ and $p_2$ on which $q$ lies, we have $s(z) < +\infty$, and then once again $s(z)\cdot z$ gives a path from $p_1$ to $p_2$ in $\partial X$.
