I'm not sure what I'm allowed to use here. Here is a proof, and I'll see whether or not you are happy with it.

Let $L$ be the lattice of those characters of $T$ which occur in finite dimensional representations of $G$. It is clearly a lattice because, if $\lambda$ occurs in $V$ and $\mu$ occurs in $U$, then $\lambda+\mu$ occurs in $V \otimes U$. As you say, it is enough to show that $L = X^*(T)$. Also, note that the root lattice is in $L$. 

Let $\lambda$ be any element of $X^*(T)$. Let $s : T \to \mathbb{C}$ be the function $t \mapsto \sum_{w \in W} \lambda(w(t))$. (Implicitly using that the Weyl group is finite; am I allowed to assume this?) Then $s$ is a $W$-invariant function on $T$, and it is not zero since it is $|W|$ at the identity. Since every conjugacy class of $G$ intersects $T$, and does so in a $W$-orbit, we can extend $s$ to a conjugacy invariant function on $G$. 

By the Peter-Weyl theorem, there must be some finite dimensional representation $V$ such that $\langle s, \chi_V \rangle \neq 0$. By the [Weyl integral formula][1], $\langle s, \chi_V \rangle$ is an integral over $T$ of a product of three factors: $s$, which is a certain finite sum of characters of $T$; $\chi_V$, which is a certain finite sum of characters in $L$, and the Weyl integrand, which is a certain sum of characters in the root lattice. 
So the product of the last two terms is a sum of characters from $L$.

In order for this integrand to be nonzero, one of the characters in $s$ must be negative a character in the root lattice, so $-w(\lambda) \in L$ for some $w\in W$. Using that $L$ is $W$-stable, this shows that $\lambda \in L$.

  [1]: http://www.math.columbia.edu/~woit/notes12.pdf