Plateau's problem for concentric circles
Visualisation figure set (will be refered to with fig.1-5)
Given two concentric circles $C_1$ and $C_2$ in $\mathbb{R}^3$ with the center points $(x_1,0,0)$ and $(x_2, 0, 0)$, let $\mathbb{S}$ be the set of all $C^0$-continuous surfaces which boundaries exactly equal $C_1 \cup C_2$ (fig.1). Let $A(S)$ denote the surface area of a surface $S$. Find $\min_{S\in\mathbb{S}}A(S)$, i.e. the surface with the minimal area among all $S\in \mathbb{S}$.
My Motivation
Most recourses on this problem assume that $S$ has to be a surface of revolution (the trivial solution of two discs for some $x_1$ and $x_2$ is excluded here) and just skip to the derivation of its formula via the Euler–Lagrange equation. Although the assumption that $S$ has to be a surface of revolution seems intuitive, I have not found a proof in the material available to me and therefore tried to come up with a proof for my self. Note that proof-writing is fairly new to me.
My Lemma
Among all $S\in \mathbb{S}$ that minimize $A(S)$ there is at least one surface of revolution.
My Proof
My approach is to construct a surface of revolution $S^*$ from any surface $S\in\mathbb{S}$, such that $A(S^*) \leq A(S)$.
Let $P_\theta$ be the xz-Plane rotated around the x-axis by the angle $\theta\in [0;\pi)$ and $\mathbb{P}_n=\{P_{k\pi / n} \;|\; k\in\mathbb{N}_0, \; k < n \}$ (the set of $n$ planes such that they divide the space into $2n$ "pie-pieces" centered around the x-axis, fig.2). Let $s_{\theta,1}$ and $s_{\theta,2}$ be the two intersection curves of a surface $S$ and $P_\theta$ and $F_{\theta,1}$ and $F_{\theta,2}$ be the two parts of $S$ that lie between $P_{k\pi/n}$ and $P_{(k+1)\pi/n}$ with $k\pi/n = \theta$ (fig.3). If we now intersect a surface $S\in\mathbb{S}$ with infinitly many planes in $\mathbb{P}_n$ for $n\to \infty$, the small surfaces $F_{\theta,1}$ and $F_{\theta,2}$ are only characterized by one of their bounding curves $s_{\theta,1}$ and $s_{\theta,2}$ respectively, because $s_{(k\pi/n),m} \to s_{((k+1)\pi/n),m}$ (fig.4). Now find $F^*\in \{F_{(k\pi/n),m} \;|\; k < n, \; m \in \{1,2\} \}$ such that the area of $F^*$ is minimal over this set. Construct a new surface $S^*$ that is an element of $\mathbb{S}$ by rotating and joining $2n$ copies of $F^*$ (fig.5). $S^*$ now has at most the same surface area than the surface $S$ we started with. Therefore, if $S$ minimizes $A(S)$, $A(S)=A(S^*)$ holds and $S^*$ also minimizes $A(S)$.
My Question
Is this a valid proof? Do there have to be more constraints on the original surface? I'm really not sure if it is at least a bit rigorous. Especially the part where I state, that every small surface $F_\theta$ is only characterized by one of its bounding curves for $n \to \infty$ gives me a headache. Do you know similar proofs or good literature on that topic?