[proofplan]
We compare both modules with the Verma module $M(\lambda)$ of highest weight $\lambda$. The universal property of $M(\lambda)$ gives $\mathfrak g$-module maps from $M(\lambda)$ onto $V$ and $W$, because the chosen vectors $v$ and $w$ satisfy the same highest-weight relations as the distinguished highest-weight vector of $M(\lambda)$. Since $V$ and $W$ are irreducible, these maps are irreducible quotients of $M(\lambda)$. The unique irreducible quotient $L(\lambda)$ of $M(\lambda)$ therefore identifies both $V$ and $W$ with $L(\lambda)$.
[/proofplan]
[step:Construct quotient maps from the Verma module to the two modules]
Let $M(\lambda)$ denote the Verma module of highest weight $\lambda$ for the Borel subalgebra $\mathfrak b$, and let $v_\lambda\in M(\lambda)$ denote its distinguished highest-weight vector. By definition, $v_\lambda$ is annihilated by $\mathfrak n^+$ and satisfies $h v_\lambda=\lambda(h)v_\lambda$ for every $h\in\mathfrak h$.
The vector $v\in V$ satisfies the hypotheses required by the [universal property of Verma modules](/theorems/9375): it is annihilated by $\mathfrak n^+$ and has $\mathfrak h$-weight $\lambda$. Hence [citetheorem:9375] gives a unique $\mathfrak g$-[module homomorphism](/page/Module%20Homomorphism) $\pi_V:M(\lambda)\to V$ such that $\pi_V(v_\lambda)=v$.
Similarly, the vector $w\in W$ is annihilated by $\mathfrak n^+$ and has $\mathfrak h$-weight $\lambda$, so [citetheorem:9375] gives a unique $\mathfrak g$-module homomorphism $\pi_W:M(\lambda)\to W$ such that $\pi_W(v_\lambda)=w$.
[guided]
The purpose of introducing $M(\lambda)$ is that it is the universal module generated by one vector satisfying exactly the highest-weight relations of weight $\lambda$. Let $v_\lambda\in M(\lambda)$ be the distinguished highest-weight vector. By definition of the Verma module, it satisfies $xv_\lambda=0$ for every $x\in\mathfrak n^+$ and $hv_\lambda=\lambda(h)v_\lambda$ for every $h\in\mathfrak h$.
Now compare this with the vector $v\in V$ from the theorem statement. The statement gives $xv=0$ for every $x\in\mathfrak n^+$ and $hv=\lambda(h)v$ for every $h\in\mathfrak h$. These are precisely the hypotheses needed to apply the universal property of Verma modules. Therefore [citetheorem:9375] produces a unique $\mathfrak g$-module homomorphism $\pi_V:M(\lambda)\to V$ with $\pi_V(v_\lambda)=v$.
The same verification applies to $w\in W$: the theorem statement gives $xw=0$ for every $x\in\mathfrak n^+$ and $hw=\lambda(h)w$ for every $h\in\mathfrak h$. Hence [citetheorem:9375] gives a unique $\mathfrak g$-module homomorphism $\pi_W:M(\lambda)\to W$ with $\pi_W(v_\lambda)=w$.
[/guided]
[/step]
[step:Show the two maps are surjective irreducible quotients]
The image $\operatorname{im}\pi_V$ is a $\mathfrak g$-submodule of $V$ because $\pi_V$ is a $\mathfrak g$-module homomorphism. Since $v=\pi_V(v_\lambda)$ and $v\ne 0$, the image is nonzero. Because $V$ is irreducible, its only $\mathfrak g$-submodules are $0$ and $V$, so $\operatorname{im}\pi_V=V$. Thus $\pi_V$ is surjective.
The same argument applies to $\pi_W$. Its image is a nonzero $\mathfrak g$-submodule of the irreducible module $W$, since $w=\pi_W(v_\lambda)\ne 0$. Hence $\operatorname{im}\pi_W=W$, so $\pi_W$ is surjective.
[/step]
[step:Identify both quotients with the unique irreducible quotient of $M(\lambda)$]
By [citetheorem:9370], the Verma module $M(\lambda)$ has a unique irreducible quotient, denoted $L(\lambda)$. Since $V$ is irreducible and $\pi_V:M(\lambda)\to V$ is surjective, $V$ is an irreducible quotient of $M(\lambda)$. Therefore there exists a $\mathfrak g$-module isomorphism
\begin{align*}
\theta_V:L(\lambda)\to V.
\end{align*}
Likewise, $W$ is irreducible and $\pi_W:M(\lambda)\to W$ is surjective, so $W$ is an irreducible quotient of $M(\lambda)$. The same uniqueness theorem gives a $\mathfrak g$-module isomorphism
\begin{align*}
\theta_W:L(\lambda)\to W.
\end{align*}
[/step]
[step:Compose the two identifications to obtain the desired isomorphism]
Since $\theta_V:L(\lambda)\to V$ is a $\mathfrak g$-module isomorphism, its inverse $\theta_V^{-1}:V\to L(\lambda)$ is also a $\mathfrak g$-module isomorphism. Define $\Theta:V\to W$ by $\Theta=\theta_W\circ\theta_V^{-1}$. A composition of $\mathfrak g$-module isomorphisms is a $\mathfrak g$-module isomorphism, so $\Theta$ is a $\mathfrak g$-module isomorphism from $V$ to $W$. Hence $V\cong W$ as $\mathfrak g$-modules.
[/step]
