[proofplan]
Define $J(\lambda)$ to be the sum of all proper $\mathfrak g$-submodules of $M(\lambda)$. The only point needing proof is that this possibly infinite sum is still proper. We prove this by showing that every proper submodule has zero intersection with the highest weight space $M(\lambda)_\lambda=\mathbb C v_\lambda$, and then using the weight-space projection argument to show the same remains true after taking the sum.
[/proofplan]
[step:Define the candidate maximal submodule as the sum of all proper submodules]
Let $\mathcal S_\lambda$ denote the set of all proper $\mathfrak g$-submodules of $M(\lambda)$. Define
\begin{align*}
J(\lambda):=\sum_{N\in\mathcal S_\lambda}N.
\end{align*}
Since a sum of $\mathfrak g$-submodules is again a $\mathfrak g$-submodule, $J(\lambda)$ is a $\mathfrak g$-submodule of $M(\lambda)$. By construction, every proper $\mathfrak g$-submodule of $M(\lambda)$ is contained in $J(\lambda)$.
[/step]
[step:Show that every submodule is a direct sum of its weight spaces]
Let $N\subset M(\lambda)$ be a $\mathfrak g$-submodule. We show that
\begin{align*}
N=\bigoplus_{\mu\in\mathfrak h^*}\bigl(N\cap M(\lambda)_\mu\bigr).
\end{align*}
The inclusion from right to left holds because each summand lies in $N$. Conversely, let $u\in N$. Since $M(\lambda)$ is a weight module, write
\begin{align*}
u=\sum_{\mu\in F}u_\mu,
\end{align*}
where $F\subset\mathfrak h^*$ is a finite set and each $u_\mu\in M(\lambda)_\mu$.
Fix $\gamma\in F$. Since the weights in $F$ are distinct linear functionals on $\mathfrak h$, there exists $h_\gamma\in\mathfrak h$ such that the complex numbers $\mu(h_\gamma)$ for $\mu\in F$ are pairwise distinct. Define the polynomial
\begin{align*}
p_\gamma(t):=\prod_{\substack{\mu\in F, \mu\ne\gamma}}\frac{t-\mu(h_\gamma)}{\gamma(h_\gamma)-\mu(h_\gamma)}\in\mathbb C[t].
\end{align*}
Then $p_\gamma(\gamma(h_\gamma))=1$ and $p_\gamma(\mu(h_\gamma))=0$ for every $\mu\in F$ with $\mu\ne\gamma$. Since $N$ is stable under the action of $\mathfrak h$, the vector $p_\gamma(h_\gamma)u$ lies in $N$. On each weight vector $u_\mu$, the element $h_\gamma$ acts by the scalar $\mu(h_\gamma)$, hence
\begin{align*}
p_\gamma(h_\gamma)u=u_\gamma.
\end{align*}
Thus $u_\gamma\in N$ for every $\gamma\in F$, proving the asserted weight-space decomposition of $N$.
[guided]
Let $N\subset M(\lambda)$ be a $\mathfrak g$-submodule. We want to prove that if a vector of $N$ is written as a finite sum of weight components, then each component is again in $N$. This is not automatic for an arbitrary subspace, so we use the action of $\mathfrak h$ to build projection operators onto the finitely many weight spaces that occur in the vector.
Take $u\in N$. Since $M(\lambda)$ is a weight module, $u$ has a finite weight expansion
\begin{align*}
u=\sum_{\mu\in F}u_\mu,
\end{align*}
where $F\subset\mathfrak h^*$ is finite and $u_\mu\in M(\lambda)_\mu$ for each $\mu\in F$. Fix one weight $\gamma\in F$. Because the elements of $F$ are distinct linear functionals on $\mathfrak h$, we can choose $h_\gamma\in\mathfrak h$ such that the scalars $\mu(h_\gamma)$ are pairwise distinct as $\mu$ ranges over $F$. This choice lets us separate the finitely many weight components by [polynomial interpolation](/page/Polynomial%20Interpolation).
Define
\begin{align*}
p_\gamma(t):=\prod_{\substack{\mu\in F, \mu\ne\gamma}}\frac{t-\mu(h_\gamma)}{\gamma(h_\gamma)-\mu(h_\gamma)}\in\mathbb C[t].
\end{align*}
The denominators are nonzero by the choice of $h_\gamma$. This polynomial satisfies $p_\gamma(\gamma(h_\gamma))=1$ and $p_\gamma(\mu(h_\gamma))=0$ for all $\mu\in F$ with $\mu\ne\gamma$.
Now view $p_\gamma(h_\gamma)$ as an element of $U(\mathfrak h)\subset U(\mathfrak g)$ acting on $M(\lambda)$. Since $N$ is a $\mathfrak g$-submodule, it is stable under every element of $U(\mathfrak g)$, so $p_\gamma(h_\gamma)u\in N$. On the weight component $u_\mu$, the element $h_\gamma$ acts by multiplication by $\mu(h_\gamma)$; therefore
\begin{align*}
p_\gamma(h_\gamma)u=\sum_{\mu\in F}p_\gamma(\mu(h_\gamma))u_\mu=u_\gamma.
\end{align*}
Thus each weight component $u_\gamma$ lies in $N$. Since $u\in N$ was arbitrary, this proves
\begin{align*}
N=\bigoplus_{\mu\in\mathfrak h^*}\bigl(N\cap M(\lambda)_\mu\bigr).
\end{align*}
[/guided]
[/step]
[step:Exclude the highest weight space from every proper submodule]
Let $N\subsetneq M(\lambda)$ be a proper $\mathfrak g$-submodule. We prove that
\begin{align*}
N\cap M(\lambda)_\lambda=0.
\end{align*}
By the PBW model of Verma modules, cited in [citetheorem:9374], the highest weight space satisfies
\begin{align*}
M(\lambda)_\lambda=\mathbb C v_\lambda,
\end{align*}
and $v_\lambda$ generates $M(\lambda)$ as a $U(\mathfrak g)$-module. If $N\cap M(\lambda)_\lambda$ contained a nonzero vector $w$, then $w=cv_\lambda$ for some $c\in\mathbb C^\times$. Hence $v_\lambda=c^{-1}w\in N$. Since $N$ is a $\mathfrak g$-submodule and $v_\lambda$ generates $M(\lambda)$, this would imply
\begin{align*}
N=M(\lambda),
\end{align*}
contradicting the assumption that $N$ is proper. Therefore $N\cap M(\lambda)_\lambda=0$.
[/step]
[step:Prove that the sum of all proper submodules is still proper]
We show that $J(\lambda)$ has zero $\lambda$-weight space. Let $u\in J(\lambda)\cap M(\lambda)_\lambda$. By definition of a sum of subspaces, there exist proper $\mathfrak g$-submodules $N_1,\dots,N_m\in\mathcal S_\lambda$ and vectors $u_i\in N_i$ such that
\begin{align*}
u=\sum_{i=1}^m u_i.
\end{align*}
By the weight-space decomposition proved above, each $u_i$ decomposes as a finite sum of vectors in $N_i\cap M(\lambda)_\mu$. Taking the $\lambda$-weight component of the displayed equality gives
\begin{align*}
u=\sum_{i=1}^m (u_i)_\lambda,
\end{align*}
where $(u_i)_\lambda\in N_i\cap M(\lambda)_\lambda$. The preceding step gives $N_i\cap M(\lambda)_\lambda=0$ for every $i$, so $(u_i)_\lambda=0$ for all $i$. Hence $u=0$.
Thus
\begin{align*}
J(\lambda)\cap M(\lambda)_\lambda=0.
\end{align*}
Since $v_\lambda\in M(\lambda)_\lambda$ is nonzero, $v_\lambda\notin J(\lambda)$, and therefore $J(\lambda)\ne M(\lambda)$. Hence $J(\lambda)$ is a proper $\mathfrak g$-submodule of $M(\lambda)$.
[/step]
[step:Conclude maximality and uniqueness]
Since $J(\lambda)$ is proper and contains every proper $\mathfrak g$-submodule of $M(\lambda)$, it is maximal among proper $\mathfrak g$-submodules. If $K\subset M(\lambda)$ is any maximal proper $\mathfrak g$-submodule, then $K\in\mathcal S_\lambda$, so $K\subset J(\lambda)$ by construction. Maximality of $K$ and properness of $J(\lambda)$ force $K=J(\lambda)$. Therefore the maximal proper submodule is unique, and it is exactly the sum of all proper submodules of $M(\lambda)$.
[/step]
