[proofplan]
The proof compares the central character on the Verma module $M(\lambda)$ with the central character on the simple highest-weight module $L(\mu)$. First we record that every central element acts on $M(\lambda)$ by the Harish-Chandra central character attached to $\lambda$, because the Verma module is generated by its highest-weight vector and central elements commute with the whole enveloping algebra. This scalar action descends to every subquotient, so a composition factor isomorphic to $L(\mu)$ has the same central character as $M(\lambda)$. The [Harish-Chandra parametrization of central characters](/theorems/9398) then converts equality of central characters into equality of shifted Weyl orbits.
[/proofplan]
[step:Identify the central character on the Verma module generated by $\lambda$]
Let $U(\mathfrak g)$ be the universal enveloping algebra of $\mathfrak g$, and let $Z(U(\mathfrak g))$ denote its center. Let
\begin{align*}
\chi_\lambda:Z(U(\mathfrak g))\to \mathbb C
\end{align*}
be the central character associated to $\lambda$ by the Harish-Chandra parametrization, using the fixed positive root system $\Phi^+$ and hence the fixed $\rho$.
Let $v_\lambda\in M(\lambda)$ denote the canonical highest-weight generator. By the defining construction of the Verma module, $M(\lambda)$ is generated by $v_\lambda$ as a left $U(\mathfrak g)$-module, and every $z\in Z(U(\mathfrak g))$ satisfies
\begin{align*}
zv_\lambda=\chi_\lambda(z)v_\lambda.
\end{align*}
Fix $z\in Z(U(\mathfrak g))$. For every $u\in U(\mathfrak g)$, centrality gives $zu=uz$, and therefore
\begin{align*}
z(uv_\lambda)=u(zv_\lambda)=u(\chi_\lambda(z)v_\lambda)=\chi_\lambda(z)uv_\lambda.
\end{align*}
Since every element of $M(\lambda)$ has the form $uv_\lambda$ for some $u\in U(\mathfrak g)$, the endomorphism of $M(\lambda)$ induced by $z$ is scalar multiplication by $\chi_\lambda(z)$.
[guided]
The point of this step is that the highest-weight vector controls the action of the center on the entire Verma module. We first name the algebraic objects involved: $U(\mathfrak g)$ is the universal enveloping algebra, and
\begin{align*}
Z(U(\mathfrak g)):=\{z\in U(\mathfrak g):zu=uz\text{ for every }u\in U(\mathfrak g)\}
\end{align*}
is its center. For the fixed choice of positive roots $\Phi^+$, the Harish-Chandra construction assigns to each $\lambda\in\mathfrak h^*$ a character
\begin{align*}
\chi_\lambda:Z(U(\mathfrak g))\to \mathbb C.
\end{align*}
Let $v_\lambda\in M(\lambda)$ be the highest-weight generator. The Verma module is generated by $v_\lambda$ as a left $U(\mathfrak g)$-module, so every element of $M(\lambda)$ can be written as $uv_\lambda$ for some $u\in U(\mathfrak g)$. By the defining central-character calculation for Verma modules,
\begin{align*}
zv_\lambda=\chi_\lambda(z)v_\lambda
\end{align*}
for every $z\in Z(U(\mathfrak g))$.
Now fix $z\in Z(U(\mathfrak g))$ and take an arbitrary element $uv_\lambda\in M(\lambda)$. Since $z$ lies in the center, it commutes with $u$. Hence
\begin{align*}
z(uv_\lambda)=u(zv_\lambda).
\end{align*}
Substituting the scalar action on the highest-weight vector gives
\begin{align*}
z(uv_\lambda)=u(\chi_\lambda(z)v_\lambda).
\end{align*}
Because $\chi_\lambda(z)\in\mathbb C$ is a scalar, this becomes
\begin{align*}
z(uv_\lambda)=\chi_\lambda(z)uv_\lambda.
\end{align*}
Thus $z$ acts on every vector of $M(\lambda)$ as the same scalar $\chi_\lambda(z)$. This is the mechanism that lets central characters pass cleanly to composition factors.
[/guided]
[/step]
[step:Pass the same central character to the composition factor $L(\mu)$]
By hypothesis, $L(\mu)$ is a composition factor of $M(\lambda)$. Thus there exist $\mathfrak g$-submodules $N\subset K\subset M(\lambda)$ such that
\begin{align*}
K/N\cong L(\mu)
\end{align*}
as $\mathfrak g$-modules.
The action of $z\in Z(U(\mathfrak g))$ on $M(\lambda)$ is scalar multiplication by $\chi_\lambda(z)$, so its restriction to $K$ is scalar multiplication by $\chi_\lambda(z)$ and it preserves $N$. Therefore the induced action of $z$ on the quotient $K/N$ is also scalar multiplication by $\chi_\lambda(z)$. Transporting this action across the isomorphism $K/N\cong L(\mu)$, every $z\in Z(U(\mathfrak g))$ acts on $L(\mu)$ by the scalar $\chi_\lambda(z)$.
[/step]
[step:Identify the same scalar action with the central character of $L(\mu)$]
By the notation in the theorem statement, $L(\mu)$ is the unique simple quotient of the Verma module $M(\mu)$; this is the existence and uniqueness result for simple quotients of Verma modules [citetheorem:9402] applied to the same triangular decomposition determined by $\Phi^+$. Applying the previous central-character calculation with $\mu$ in place of $\lambda$, every $z\in Z(U(\mathfrak g))$ acts on $M(\mu)$ by $\chi_\mu(z)$, and hence acts on its quotient $L(\mu)$ by $\chi_\mu(z)$.
From the preceding step, the same element $z$ acts on $L(\mu)$ by $\chi_\lambda(z)$. Therefore
\begin{align*}
\chi_\lambda(z)=\chi_\mu(z)
\end{align*}
for every $z\in Z(U(\mathfrak g))$. Hence
\begin{align*}
\chi_\lambda=\chi_\mu
\end{align*}
as characters $Z(U(\mathfrak g))\to\mathbb C$.
[/step]
[step:Apply the Harish-Chandra parametrization and rewrite the shifted orbit condition]
The hypotheses of the Harish-Chandra parametrization of central characters [citetheorem:9398] are satisfied: $\mathfrak g$ is finite-dimensional complex semisimple, $\mathfrak h\subset\mathfrak g$ is a Cartan subalgebra, $\Phi^+$ is the fixed positive root system, $W$ is the corresponding Weyl group, and
\begin{align*}
\rho=\frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha.
\end{align*}
Using these same choices, the equality $\chi_\lambda=\chi_\mu$ holds if and only if $\lambda+\rho$ and $\mu+\rho$ lie in the same $W$-orbit. Therefore there exists $w\in W$ such that
\begin{align*}
\mu+\rho=w(\lambda+\rho).
\end{align*}
Subtracting $\rho$ from both sides gives
\begin{align*}
\mu=w(\lambda+\rho)-\rho.
\end{align*}
This is precisely $\mu=w\cdot\lambda$ for the dot action $w\cdot\lambda:=w(\lambda+\rho)-\rho$, so $\mu$ lies in the $W$-dot orbit of $\lambda$.
[/step]
