[proofplan]
The proof reduces concentration of a Lipschitz function to concentration of neighbourhoods of sets of measure at least one half. The median gives two large sets, the sublevel set $\{f \leq m_f\}$ and the superlevel set $\{f \geq m_f\}$. Lipschitz continuity forces the upper tail and lower tail to lie outside Euclidean neighbourhoods of these large sets. The spherical isoperimetric concentration inequality then bounds both tails, and the union bound gives the stated estimate.
[/proofplan]
[step:Record the spherical concentration input for sets of measure at least one half]
We use the following standard consequence of the spherical isoperimetric inequality: there is a universal constant $a>0$ such that, for every $n \geq 2$, every Borel set $A \subset S^{n-1}$ with $\sigma_{n-1}(A) \geq 1/2$, and every $r>0$,
\begin{align*}
\sigma_{n-1}\bigl(S^{n-1} \setminus A_r\bigr)
\leq \exp(-a n r^2),
\end{align*}
where the Euclidean $r$-neighbourhood of $A$ in $S^{n-1}$ is
\begin{align*}
A_r := \{x \in S^{n-1}: \operatorname{dist}(x,A) < r\},
\end{align*}
and
\begin{align*}
\operatorname{dist}(x,A) := \inf_{y \in A}|x-y|.
\end{align*}
(citing a result not yet in the wiki: Spherical Isoperimetric Inequality)
[/step]
[step:Bound the upper tail by the complement of a neighbourhood of the lower median set]
Define the Borel set
\begin{align*}
A_- := \{y \in S^{n-1}: f(y) \leq m_f\}.
\end{align*}
Since $m_f$ is a median, $\sigma_{n-1}(A_-) \geq 1/2$. Let $t>0$ and define $r := t/L$. If $x \in S^{n-1}$ satisfies $f(x) \geq m_f+t$, then for every $y \in A_-$,
\begin{align*}
t \leq f(x)-m_f \leq f(x)-f(y) \leq |f(x)-f(y)| \leq L|x-y|.
\end{align*}
Taking the infimum over $y \in A_-$ gives $\operatorname{dist}(x,A_-) \geq t/L = r$. Therefore
\begin{align*}
\{x \in S^{n-1}: f(x) \geq m_f+t\}
\subseteq S^{n-1}\setminus (A_-)_r.
\end{align*}
Applying the spherical concentration input to $A_-$ gives
\begin{align*}
\sigma_{n-1}(\{x \in S^{n-1}: f(x) \geq m_f+t\})
\leq \exp\left(-\frac{a n t^2}{L^2}\right).
\end{align*}
[guided]
The median gives a set of measure at least one half on which $f$ lies below the median. Define
\begin{align*}
A_- := \{y \in S^{n-1}: f(y) \leq m_f\}.
\end{align*}
This set is Borel because $f$ is Lipschitz, hence continuous, and $\sigma_{n-1}(A_-) \geq 1/2$ by the definition of median.
The key point is that a point in the upper tail cannot be close to $A_-$. Let $t>0$ and set $r := t/L$. If $x \in S^{n-1}$ satisfies $f(x) \geq m_f+t$, then for every $y \in A_-$ we have $f(y) \leq m_f$, hence
\begin{align*}
t \leq f(x)-m_f \leq f(x)-f(y) \leq |f(x)-f(y)| \leq L|x-y|.
\end{align*}
The last inequality is exactly the $L$-Lipschitz hypothesis. Since this inequality holds for every $y \in A_-$, taking the infimum over $A_-$ yields
\begin{align*}
\operatorname{dist}(x,A_-) \geq \frac{t}{L}=r.
\end{align*}
Thus $x$ is not in the Euclidean $r$-neighbourhood $(A_-)_r$, so
\begin{align*}
\{x \in S^{n-1}: f(x) \geq m_f+t\}
\subseteq S^{n-1}\setminus (A_-)_r.
\end{align*}
Now the spherical concentration input applies because $\sigma_{n-1}(A_-) \geq 1/2$. Therefore
\begin{align*}
\sigma_{n-1}(\{x \in S^{n-1}: f(x) \geq m_f+t\})
\leq \sigma_{n-1}\bigl(S^{n-1}\setminus (A_-)_r\bigr)
\leq \exp(-a n r^2)
= \exp\left(-\frac{a n t^2}{L^2}\right).
\end{align*}
[/guided]
[/step]
[step:Bound the lower tail by the complement of a neighbourhood of the upper median set]
Define the Borel set
\begin{align*}
A_+ := \{y \in S^{n-1}: f(y) \geq m_f\}.
\end{align*}
Since $m_f$ is a median, $\sigma_{n-1}(A_+) \geq 1/2$. Let $t>0$ and set $r := t/L$. If $x \in S^{n-1}$ satisfies $f(x) \leq m_f-t$, then for every $y \in A_+$,
\begin{align*}
t \leq m_f-f(x) \leq f(y)-f(x) \leq |f(y)-f(x)| \leq L|x-y|.
\end{align*}
Taking the infimum over $y \in A_+$ gives $\operatorname{dist}(x,A_+) \geq r$. Hence
\begin{align*}
\{x \in S^{n-1}: f(x) \leq m_f-t\}
\subseteq S^{n-1}\setminus (A_+)_r.
\end{align*}
Applying the spherical concentration input to $A_+$ gives
\begin{align*}
\sigma_{n-1}(\{x \in S^{n-1}: f(x) \leq m_f-t\})
\leq \exp\left(-\frac{a n t^2}{L^2}\right).
\end{align*}
[/step]
[step:Combine the one-sided estimates by the union bound]
The two-sided deviation event decomposes as
\begin{align*}
\{x \in S^{n-1}: |f(x)-m_f| \geq t\}
\subseteq
\{x \in S^{n-1}: f(x) \geq m_f+t\}
\cup
\{x \in S^{n-1}: f(x) \leq m_f-t\}.
\end{align*}
Using finite subadditivity of the probability measure $\sigma_{n-1}$ and the two one-sided bounds, we obtain
\begin{align*}
\sigma_{n-1}(\{x \in S^{n-1}: |f(x)-m_f| \geq t\})
&\leq
\sigma_{n-1}(\{x \in S^{n-1}: f(x) \geq m_f+t\}) \\
&\quad+
\sigma_{n-1}(\{x \in S^{n-1}: f(x) \leq m_f-t\}) \\
&\leq 2\exp\left(-\frac{a n t^2}{L^2}\right).
\end{align*}
Thus the theorem holds with the universal constants $C:=2$ and $c:=a$.
[/step]
