Here's maybe another (more conceptual?) way to think about it.

First of all, if $\mu_1$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_1}$, and $\mu_2$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_2}$, then $\mu_1+\mu_2$ is a dominant weight which appears with nonzero multiplicity in $V^{\lambda_1+\lambda_2}$. (This is just saying that if $\lambda_1-\mu_1$ is a nonnegative sum of simple roots, and ditto for $\lambda_2-\mu_2$, then definitely $(\lambda_1+\lambda_2)-(\mu_1-\mu_2)$ is as well.)

The previous paragraph reduces the problem to showing that if $\lambda=\omega_i$ is a fundamental weight, then for all $m\gg0$ sufficiently large, $V^{m\omega_i}$ contains a weight strictly in the fundamental chamber. Let's show this.

Recall the Weyl vector $\rho:=\frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha$, which is well known to satisfy $\rho= \sum_{j=1}^{n}\omega_j$. Let $\Phi_i$ denote the maximal parabolic root system generated by the simple roots $\alpha_j$ for $j\neq i$, and let's use the notation $\rho_i :=\frac{1}{2}\sum_{\alpha\in\Phi_i^+}\alpha$ for its "parabolic Weyl vector." Note that $\langle \rho_i,\alpha^\vee_j\rangle = 1$ for all $j\neq i$ precisely because this is the Weyl vector of the parabolic root system, while $\langle \rho_i, \alpha^\vee_i \rangle < 0$ since no factor of $\alpha_i$ appears at all in $\rho_i$ but some $\alpha_j$ for $j$ a node adjacent to $i$ in the Dynkin diagram does appear.

Therefore $2\rho - 2\rho_i = \sum_{\alpha\in \Phi^+\setminus \Phi_i^+}\alpha = \sum_{j\neq i}-2\omega_j + c\omega_i$ for some $c > 0$. So we have $2\rho + (2\rho-2\rho_i)=(c+2)\omega_i$, where $(2\rho-2\rho_i)$ is a nonnegative sum of simple roots, which is to say the strictly dominant weight $2\rho$ appears in $V^{(c+2)\omega_i}$. And then for any $m\geq(c+2)$, the strictly dominant weight $2\rho + (m-(c+2))\omega_i$ appears in $V^{m\omega_i}$, completing the proof.

Here's maybe another (more conceptual?) way to think about it.

First of all, if $\mu_1$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_1}$, and $\mu_2$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_2}$, then $\mu_1+\mu_2$ is a dominant weight which appears with nonzero multiplicity in $V^{\lambda_1+\lambda_2}$. (This is just saying that if $\lambda_1-\mu_1$ is a nonnegative sum of simple roots, and ditto for $\lambda_2-\mu_2$, then definitely $(\lambda_1+\lambda_2)-(\mu_1-\mu_2)$ is as well.)

The previous paragraph reduces the problem to showing that if $\lambda=\omega_i$ is a fundamental weight, then for all $m\gg0$ sufficiently large, $V^{m\omega_i}$ contains a weight strictly in the fundamental chamber. Let's show this.

Recall the Weyl vector $\rho:=\frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha$, which is well known to satisfy $\rho= \sum_{j=1}^{n}\omega_j$. Let $\Phi_i$ denote the maximal parabolic root system generated by the simple roots $\alpha_j$ for $j\neq i$, and let's use the notation $\rho_i :=\frac{1}{2}\sum_{\alpha\in\Phi_i^+}\alpha$ for its "parabolic Weyl vector." Note that $\langle \rho_i,\alpha^\vee_j\rangle = 1$ for all $j\neq i$ precisely because this is the Weyl vector of the parabolic root system, while $\langle \rho_i, \alpha^\vee_i \rangle < 0$ since no factor of $\alpha_i$ appears at all in $\rho_i$ but some $\alpha_j$ for $j$ a node adjacent to $i$ in the Dynkin diagram does appear.

Therefore $2\rho - 2\rho_i = \sum_{\alpha\in \Phi^+\setminus \Phi_i^+}\alpha= c\omega_i$ for some integer $c > 0$. But one of the $\alpha \in \Phi^+\setminus \Phi_i^+$ is the highest root $\theta$ of $\Phi$, which when written $\theta=\sum_{j=1}^{n}a_j\alpha_j$ satisfies $a_j \geq 1$. So there is some multiple $k$ of $2\rho - 2\rho_i$ for which $k(2\rho - 2\rho_i) - 2\rho$ is a nonnegative sum of simple roots. This is to say the strictly dominant weight $2\rho$ appears in $V^{kc\omega_i}$. And then for any $m\geq kc$, the strictly dominant weight $2\rho + (m-kc)\omega_i$ appears in $V^{m\omega_i}$, completing the proof.

Here's maybe another (more conceptual?) way to think about it.

First of all, if $\mu_1$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_1}$, and $\mu_2$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_2}$, then $\mu_1+\mu_2$ is a dominant weight which appears with nonzero multiplicity in $V^{\lambda_1+\lambda_2}$. (This is just saying that if $\lambda_1-\mu_1$ is a nonnegative sum of simple roots, and ditto for $\lambda_2-\mu_2$, then definitely $(\lambda_1+\lambda_2)-(\mu_1-\mu_2)$ is as well.)

The previous paragraph reduces the problem to showing that if $\lambda=\omega_i$ is a fundamental weight, then for all $m\gg0$ sufficiently large, $V^{m\omega_i}$ contains a weight strictly in the fundamental chamber. Let's show this.

Recall the Weyl vector $\rho:=\frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha$, which is well known to satisfy $\rho= \sum_{j=1}^{n}\omega_j$. Let $\Phi_i$ denote the maximal parabolic root system generated by the simple roots $\alpha_j$ for $j\neq i$, and let's use the notation $\rho_i :=\frac{1}{2}\sum_{\alpha\in\Phi_i^+}\alpha$ to for its "parabolic Weyl vector." Note that $\langle \rho_i,\alpha^\vee_j\rangle = 1$ for all $j\neq i$ precisely because this is the Weyl vector of the parabolic root system, while $\langle \rho_i, \alpha^\vee_i \rangle < 0$ since no factor of $\alpha_i$ appears at all in $\rho_i$ but some $\alpha_j$ for $j$ a node adjacent to $i$ in the Dynkin diagram does appear.

Therefore $2\rho - 2\rho_i = \sum_{\alpha\in \Phi^+\setminus \Phi_i^+}\alpha = \sum_{j\neq i}-2\omega_j + c\omega_i$ for some $c > 0$. So we have $2\rho + (2\rho-2\rho_i)=(c+2)\omega_i$, where $(2\rho-2\rho_i)$ is a nonnegative sum of simple roots, which is to say the strictly dominant weight $2\rho$ appears in $V^{(c+2)\omega_i}$. And then for any $m\geq(c+2)$, the strictly dominant weight $2\rho + (m-(c+2))\omega_i$ appears in $V^{m\omega_i}$, completing the proof.

Here's maybe another (more conceptual?) way to think about it.

First of all, if $\mu_1$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_1}$, and $\mu_2$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_2}$, then $\mu_1+\mu_2$ is a dominant weight which appears with nonzero multiplicity in $V^{\lambda_1+\lambda_2}$. (This is just saying that if $\lambda_1-\mu_1$ is a nonnegative sum of simple roots, and ditto for $\lambda_2-\mu_2$, then definitely $(\lambda_1+\lambda_2)-(\mu_1-\mu_2)$ is as well.)

The previous paragraph reduces the problem to showing that if $\lambda=\omega_i$ is a fundamental weight, then for all $m\gg0$ sufficiently large, $V^{m\omega_i}$ contains a weight strictly in the fundamental chamber. Let's show this.

Recall the Weyl vector $\rho:=\frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha$, which is well known to satisfy $\rho= \sum_{j=1}^{n}\omega_j$. Let $\Phi_i$ denote the maximal parabolic root system generated by the simple roots $\alpha_j$ for $j\neq i$, and let's use the notation $\rho_i :=\frac{1}{2}\sum_{\alpha\in\Phi_i^+}\alpha$ for its "parabolic Weyl vector." Note that $\langle \rho_i,\alpha^\vee_j\rangle = 1$ for all $j\neq i$ precisely because this is the Weyl vector of the parabolic root system, while $\langle \rho_i, \alpha^\vee_i \rangle < 0$ since no factor of $\alpha_i$ appears at all in $\rho_i$ but some $\alpha_j$ for $j$ a node adjacent to $i$ in the Dynkin diagram does appear.

Therefore $2\rho - 2\rho_i = \sum_{\alpha\in \Phi^+\setminus \Phi_i^+}\alpha = \sum_{j\neq i}-2\omega_j + c\omega_i$ for some $c > 0$. So we have $2\rho + (2\rho-2\rho_i)=(c+2)\omega_i$, where $(2\rho-2\rho_i)$ is a nonnegative sum of simple roots, which is to say the strictly dominant weight $2\rho$ appears in $V^{(c+2)\omega_i}$. And then for any $m\geq(c+2)$, the strictly dominant weight $2\rho + (m-(c+2))\omega_i$ appears in $V^{m\omega_i}$, completing the proof.