"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 $$\exp 2 \pi i x = 1 \, \implies \, x \in \mathbb Z.$$

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 $$ L = \{ \xi \in \mathfrak t_0 \mid \exp \lambda = 1 \} $$ 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.

"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 $$\exp 2 \pi i x = 1 \, \implies \, x \in \mathbb Z.$$

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.

"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 $$\exp 2 \pi i x = 1 \, \implies \, x \in \mathbb Z.$$

"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 $$\exp 2 \pi i x = 1 \, \implies \, x \in \mathbb Z.$$

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 $$ L = \{ \xi \in \mathfrak t_0 \mid \exp \lambda = 1 \} $$ 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.