Let $G$ be a compact connected Lie group, $T$ maximal torus, $X^*(T)$

the set of characters of $T$, identified somehow with $\mathbb{Z}^n$. Let next $\Phi$ denote the set of roots, corresponding to $T$, $$ \Lambda=\{v\in \mathbb{R}^n: \forall \alpha\in \Phi,\ 2(v,\alpha)/(\alpha,\alpha)\in \mathbb{Z}\}$$ be the lattice of weights. Then $X^*(T)\subset \Lambda$.

What is the easiest and most elementary way to prove it?

It suffices to prove that for any $\lambda\in X^*(T)$ is such a character that for some finite-dimensional complex representation of $G$ there is a non-zero $T$-invariant subspace, on which $T$ acts by multiplying by $\lambda(\cdot)$. Is there any way to do it without appealing to infinite-dimensional representations (inducing $\lambda$ from $T$ to $G$ and some work after that)?

Let $G$ be a compact connected Lie group, $T$ maximal torus, identified with $\mathbb{R}^n/\mathbb{Z}^n$, $X^*(T)$

the set of characters of $T$, naturally identified with $\mathbb{Z}^n$. Let next $\Phi$ denote the set of roots, corresponding to $T$, $$ \Lambda=\{v\in \mathbb{R}^n: \forall \alpha\in \Phi,\ 2(v,\alpha)/(\alpha,\alpha)\in \mathbb{Z}\}$$ be the lattice of weights. Then $X^*(T)\subset \Lambda$.

What is the easiest and most elementary way to prove it?

It suffices to prove that for any $\lambda\in X^*(T)$ is such a character that for some finite-dimensional complex representation of $G$ there is a non-zero $T$-invariant subspace, on which $T$ acts by multiplying by $\lambda(\cdot)$. Is there any way to do it without appealing to infinite-dimensional representations (inducing $\lambda$ from $T$ to $G$ and some work after that)?

Let $G$ be a compact connected Lie group, $T$ maximal torus, $X^*(T)$ the set of characters of $T$, identified somehow with $\mathbb{Z}^r$. Let next $\Phi$ denote the set of roots, corresponding to $T$, $$ \Lambda=\{v\in \mathbb{R}^n: \forall \alpha\in \Phi\, 2(v,\alpha)/(\alpha,\alpha)\in \mathbb{Z}\} $$ be the lattice of weights. Then $X^*(T)\subset \Lambda$. What is the easiest and most elementary way to prove it? It suffices to prove that for any $\lambda\in X^*(T)$ is such a character that for some finite-dimensional complex representation of $G$ there is a non-zero $T$-invariant subspace, on which $T$ acts by multiplying by $\lambda(\cdot)$. Is there any way to do it without appealing to infinite-dimensional representations (inducing $\lambda$ from $T$ to $G$ and some work after that)?

Let $G$ be a compact connected Lie group, $T$ maximal torus, $X^*(T)$ 

the set of characters of $T$, identified somehow with $\mathbb{Z}^n$. Let next $\Phi$ denote the set of roots, corresponding to $T$, $$ \Lambda=\{v\in \mathbb{R}^n: \forall \alpha\in \Phi,\ 2(v,\alpha)/(\alpha,\alpha)\in \mathbb{Z}\}$$ be the lattice of weights. Then $X^*(T)\subset \Lambda$.

What is the easiest and most elementary way to prove it?

It suffices to prove that for any $\lambda\in X^*(T)$ is such a character that for some finite-dimensional complex representation of $G$ there is a non-zero $T$-invariant subspace, on which $T$ acts by multiplying by $\lambda(\cdot)$. Is there any way to do it without appealing to infinite-dimensional representations (inducing $\lambda$ from $T$ to $G$ and some work after that)?