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)?



why are all characters of the maximal torus in a Lie group weights? 
"Easiest" depends on how you set things up: everything really hinges on how you want to identify $X^\ast(T)$ with $\mathbb Z^n$. It's probably cleanest if you don't work explicitly with $\mathbb Z^n$, but instead state everything in terms of Lie algebras and their duals. I personally like the setup given in Knapp, *Lie Groups Beyond an Introduction*, IV.6--7, which is fairly standard. In the end it all boils down to associating a copy of $SU(2)$ (or $\mathfrak{su}(2)$ or $\mathfrak{sl}\_2(\mathbb C)$ ...) to each root $\alpha$, and then the integrality statement you're after ultimately follows from the fact that $$\exp 2 \pi i x = 1 \, \implies \, x \in \mathbb Z.$$




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**Edit:** Here's an outline of the details. Let $\mathfrak t\_0$ and $\mathfrak t$ denote, respectively, the real and complex Lie algebras of $T$ and set
$$ L = \{ \xi \in \mathfrak t\_0 \mid \exp \xi = 1 \} $$
for the kernel of the exponential map $\exp \colon \mathfrak t\_0 \to T$.
Also define
$$ L^\perp = \{ \lambda \in \mathfrak t^\ast \mid \lambda(L) \subset 2\pi i \mathbb Z \}. $$
(Admittedly, the $^\perp$ is a slight abuse of notation.) Then there is an isomorphism
$$ L^\perp \stackrel{\sim}{\longrightarrow} X^\ast(T) $$
given by sending $\lambda$ to the character $e^\lambda$ defined by
$$ e^\lambda(\exp \xi) = e^{\lambda(\xi)}, \qquad \xi \in \mathfrak t\_0. $$
This is basically Proposition 4.58 in Knapp. So our objective now is to show that $L^\perp$ sits in the weight lattice $\Lambda$ --- in other words, we want to show that
$$ 2\frac{(\lambda, \alpha)}{(\alpha,\alpha)} \in \mathbb Z $$
for all $\lambda \in L^\perp$ and $\alpha \in \Phi$ (Prop 4.59). To this end, let $\psi\_\alpha \colon SU(2) \to G$ denote the root morphism corresponding to the root $\alpha \in \Phi$. This morphism has the property that $d \psi\_\alpha(h) = 2 \alpha^\vee/(\alpha,\alpha)$, where
$$ h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \in \mathfrak{su}(2). $$
Consequently,
$$ 1 = \psi\_\alpha(1) = \psi\_\alpha(\exp (2\pi i h)) = \exp (2 \pi i \, d \psi\_\alpha(h)), $$
that is,
$$ 2 \pi i \, d \psi\_\alpha(h) = 2 \pi i \cdot 2 \frac{\alpha^\vee}{(\alpha,\alpha)} \in L $$
whence
$$ \lambda(2 \pi i \cdot 2 \frac{\alpha^\vee}{(\alpha,\alpha)}) \in 2 \pi i \mathbb Z \iff 2\frac{(\lambda, \alpha)}{(\alpha,\alpha)} \in \mathbb Z, $$
as desired.


