Homomorphisms from a simple Verma module have dimension at most one

Suppose that $f,g:M(\mu )\to M(\lambda )$ are linearly independent. They are injective. Every endomorphism of $M(\mu )$ is scalar: it preserves the one-dimensional highest-weight space (visible in the character formula), and that highest vector generates the module. If the two simple images meet nontrivially, their intersection equals both images, so $f^{-1}g$ is such an endomorphism, contradicting independence. Thus $f(M(\mu ))\oplus g(M(\mu ))$ embeds in $M(\lambda )$.

Write $\delta =\lambda -\mu \in Q^{+}$, since a nonzero map sends the highest vector to a target weight vector, and put $d=\mathrm{ht}(\delta )$. List the positive roots as ${\alpha }_1,\dots ,{\alpha }_m$ and set $h_j=\mathrm{ht}({\alpha }_j)>0$. By the character formula, the sum of weight-space dimensions at heights at most $N$ below any Verma highest weight is $D(N)=\mathrm{\#}\{(k_1,\dots ,k_m)\in {\mathbb{Z}}_{\ge 0}^m:{\sum }_j{h}_j{k}_j\le N\}\mathrm{.}$ Both copies of the source's height-at-most-$N$ subspace map into the target's height-at-most-$N+d$ subspace, giving $2D(N)\le D(N+d)$. To compare these counts, let $S_t=\{x\in {\mathbb{R}}_{\ge 0}^m:{\sum }_j{h}_j{x}_j\le t\}$ and $H={\sum }_j{h}_j$. The union of unit cubes based at the integer points counted by $D(N)$ contains $S_N$ and is contained in $S_{N+H}$, up to boundaries of volume zero. Since $\mathrm{vol}(S_t)=t^m\mathrm{/}(m!{\prod }_j{h}_j)$, this yields $D(N)\sim N^m\mathrm{/}(m!{\prod }_j{h}_j)$ and hence $D(N+d)\mathrm{/}D(N)\to 1$. If $m=0$, both counts are instead exactly $1$. In either case $2D(N)\le D(N+d)$ is impossible for large $N$.