Every Verma module contains a simple Verma submodule

Suppose no embedded Verma submodule of $M(\lambda )$ is simple. Any proper nonzero submodule of an embedded Verma contains a singular vector; its generated submodule is another embedded Verma by injectivity. A proper inclusion strictly lowers the highest weight, so successive such submodules have weights $\lambda -{\beta }_j$ with strictly increasing heights of ${\beta }_j\in Q^{+}$.

Every such embedded Verma has the same Casimir scalar as $M(\lambda )$, so its nonzero ${\beta }_j$ satisfies $2(\lambda +\rho ,{\beta }_j)=({\beta }_j,{\beta }_j)$. Taking real parts in the real span of the roots places ${\beta }_j$ on a bounded sphere centred at the real part of $\lambda +\rho$. The root lattice is discrete, so the set $F$ of its solutions is finite.

Let $m=\mathrm{\mid }F\mathrm{\mid }$. Starting with $M(\lambda )$, apply step 1.1 only $m+1$ times. Finite induction and finite choice suffice to obtain $m+1$ proper, strictly nested embedded Vermas. Their distinct nonzero weights give $m+1$ distinct members of $F$, a contradiction. Hence some embedded Verma is simple.