Let $S$ be a closed convex set of $\mathbb{R}^2.$ Let us assume that $S$ is NOT a strip (which has a boundary with two connected components.) How can I prove that $\partial S$ is connected ? It seems pretty evident on a picture but the complete formal proof is quite difficult for me.
Thanks for any help.
Edit I've just found this article where the proof is given in $\mathbb{R}^3$. I think it can be adapted in $\mathbb{R}^2$. However I am still open for straightforward arguments for plane convex sets.
Here is a sequence of lemmas leading to the proof. Let me know which of these lemmas you cannot prove.
The claim is clear if $S$ has empty interior (in this case, $S=\partial S$). Hence, assume that $int(S)\ne \emptyset$. Pick a point $o\in int(S)$ and a small circle $C$ centered at $o$ and contained in $S$. I will identify $o$ with the origin in $R^2$. Let $p: \partial S\to C$ be the radial projection.
Lemma 1. $p$ is a homeomorphism to its image. (Hint: Show first that $p$ is continuous and injective; then use the fact that $\partial S$ and $C$ are 1-dimensional manifolds.)
Corollary 1. If the image of $p$ is connected, so is $\partial S$.
Lemma 2. If $I:=p(\partial S)$ is disconnected and $a, b\in C\setminus I$ are points separating the image, then $a, b$ are antipodal on $C$ (centrally symmetric about $o$), $a=-b$.
Corollary 2. If $I$ is disconnected, then $I= C \setminus \{a, -a\}$.
In this situation, let $L$ denote the linear span of $\{a\}$, i.e. the line through $a, -a$.
Lemma 3. If $I$ is disconnected then $S$ is a parallel strip, parallel to the line $L$. (Hint: Using the fact that $L\subset S$, show that with every point $x\in S$, the set $S$ contains the line $L_x$ parallel to $L$.)