Any bounded region $D\subset\mathbb C^n$ with spherical symmetry is a ball or a ball-shell 
Theorem. Any bounded region $D\subset\mathbb C^n$ with spherical symmetry is either a ball or ball-shell centered at the origin.

Notation:

*

*A ball in $\mathbb C^n$ is a set of the form $$B^n_r(a) = \{z\in \Bbb C^n: \|z-a\|_2 < r\}$$

*A ball-shell in $\Bbb C^n$ is a set of the form $$A^n(a,r,R) = \{z\in \mathbb C^n: r < \|z-a\|_2 < R\}$$

*A region is an open, connected subset of $\mathbb C^n$.

By definition,

An open set $D\subset \mathbb C^n$ has spherical symmetry if for all unitary matrices $U:\Bbb C^n\to\Bbb C^n$, $UD\subset D$.

The argument given to me proceeds as follows.

Let $r := \inf_{z\in D} \|z\|_2$ and $R := \sup_{z\in D} \|z\|_2$. If $0\in D$, $D\subset B^n_R(0)$. If $0\notin D$, $D\subset A^n(0,r,R)$. Due to the spherical symmetry of $D$, we have $$D = \bigcup_{z\in D} \partial B^n_{\|z\|_2} (0)$$
Since $\{\|z\|_2: z\in D\}$ is a union of finitely many relatively open intervals in $\Bbb R_+$, we are done.

I understand everything except the last line and how the conclusion follows. $z\mapsto \|z\|_2$ is a continuous function, and $D$ is connected, so $\{\|z\|_2: z\in D\} \subset \Bbb R_+$ must be connected. The connected subsets of $\mathbb R$ are exactly intervals and points. What is next? How do we conclude that any bounded region $D\subset\mathbb C^n$ with spherical symmetry is either a ball or ball-shell? Thanks!
 A: This answer is collected from the comments.
You argued correct that $I:= \{\|z\|_2 \colon z \in D\}$ is a connected subset of $\mathbb{R}_+$, hence a point or an interval.
We show that $\sup I$ is not an element of $I$, so that $I$ is open at its right endpoint.
If $\sup I$ was an element of $I$, then there existed $z \in D$ with $\|z\|_2 = \sup I$.
But $D$ is open, so $D$ contains an open ball around $z$ and $I$ therefore contains elements larger than $\sup I$, a contradiction.
The argument for the left endpoint of $I$ is analogous except in the case that $0 \in D$, in which case $I = [0,\sup I)$ is still open in the relative topology of $\mathbb{R}_+$.
Knowing that $I$ is an open connected interval, it follows that
$$\bigcup_{z \in D} \partial B^n_{\|z\|_2}(0) = \bigcup_{s \in I} \partial B^n_s(0) = \begin{cases} A^n(0,r,R) &\text{if } 0 \notin D \text{ so that } I = (r,R), \\ D^n_R(0) &\text{if } 0 \in D \text{ so that } I = [0,R). \end{cases}$$
Just for the sake of it, I will expand the alternative argument to reach the final conclusion, which is sketched in the comments.
I treat the case $0 \notin D$, the other case is analogous.
Take $z,w \in \bar{D}$ with $\|z\|_2=r$ and $\|w\|_2=R$. NB: we already know that $w,z \in \partial D$ since $I$ is open.
Claim: there exists a path $\gamma \colon [0,1] \rightarrow \bar{D}$ from $z$ to $w$ with $\gamma((0,1)) \subset D$.
If the claim holds, then by continuity of $\gamma$, for any $s \in (r,R)$, there is a point on the path with norm $s$.
By spherical symmetry, it follows that $D$ contains all points of norm $s$.
Since this holds for any $s \in (r,R)$, we found that $D = A^n(0,r,R)$.
Proof of claim:
First note that $D$ is path-connected, because it is an open connected subset of $\mathbb{C}^n$.
Since $z \in \bar{D}$, there exists a sequence $(z_n) \subset D$ converging to $z$.
Since $D$ is path-connected, we can take paths $\gamma_n \colon [\frac{1}{n+1},\frac{1}{n}] \rightarrow D$ joining $z_n$ and $z_{n+1}$.
Define $\gamma \colon (0,1/2) \rightarrow D$ by $\gamma(t) = \gamma_n(t)$ for $t \in [\frac{1}{n+1},\frac{1}{n}]$.
Then $\gamma(1/n) = z_n$, so $\lim_{n \rightarrow \infty} \gamma(1/n) = z$ exists and we can continuously extend$^{(1)}$ $\gamma$ to a function $[0,1/2) \rightarrow D$.
Do the same with a sequence $(w_n) \subset D$ converging to $w$ to define $\gamma$ on $(1/2,1)$. Then join $z_2$ and $w_2$ and insert this final path-segment in the middle of $\gamma$.
$^{(1)}$ Actually, we need that $\lim_{n \rightarrow \infty} \gamma(x_n) = z$ for any sequence $x_n$ converging to $z$. To achieve this, pick the initial sequence $(z_n)$ and the paths $\gamma_n$ with some more care, e.g. so that $\gamma_n$ is always a straight line form $z_n$ to $z_{n+1}$.
