Possibly this argument works; I am also new in Lie groups and Lie algebras, so I am happy for experts comments.

Consider $\lambda$ a minuscule i.e. a minimal dominant weight (according to Humphreys or Bourbaki).

**1.** $\lambda + m_1\alpha_1 + \cdots  m_l\alpha_l $, for $m_i\geq 0$, is *not* minimal dominant weight unless all $m_i=0$.

Here all $m_i\geq 0$. If some $m_i>0$ then $\lambda'=\lambda + m_1\alpha_1 + \cdots  m_l\alpha_l $ is bigger than $\lambda$ (i.e. by def.,$\lambda'-\lambda$ is sum of fundamental roots).  Then minimality of both $\lambda$ and $\lambda'$ (as dominant weights) implies that they should be equal so $m_1\alpha_1 + \cdots  m_l\alpha_l =0$; but $\alpha_i$'s are independent and one $m_i$ is non-zero, so contradiction.

 **2.** Every dominant weight can be written as $\sum q_i\alpha_i$ where $q_i$ are positive rationals.

If $\lambda$ is dominant weight then this means $\lambda=c_1\lambda_1 + \cdots + c_l\lambda_l$ where $c_i\in\mathbb{Z}^+$. Further, each $\lambda_i$ can be written as $a_{i1}\alpha_1 + \cdot + a_{il}\alpha_l$ where $a_{ij}$ are positive rationals (see Exercise 13.8 in Humphreys Lie algebra). This proves the assertion (2).

 **3** (Main fact:) *Every minimal dominant weight $\lambda$ can be written as $\sum q_i\alpha_i$ where $q_i$'s are positive rationals, less than $1$.*

Write a dominant weight $\lambda$ as $\sum q_i'\alpha_i$ with $q_i'$ positive rationals. If some $q_j'>1$ then we can subtract $\alpha_j$ from $\lambda$ to get a smaller dominant weight.

**4.** If $\lambda,\nu$ are minimal dominant weights in same coset of $\Lambda_r$, then $\lambda-\mu$ is integral combination of $\alpha_i$'s; so
$$\lambda=\nu + c_1\alpha_1 + \cdots + c_l\alpha_l, (c_i\in\mathbb{Z}).$$

If all $c_i's$ are $\geq 0$, apply $1$; if all $c_i\leq 0$, then take them to left side and apply 1.

If some $c_i>0$ then in RHS, coefficient of $\alpha_i$ will be rational $>1$, but in LHS, it is less than $1$, contradiction to (3).