Suppose $G$ is a compact abelian group, and suppose $<g^n>$ is dense in $G$ where $g$ is a particular element of $G$ and $<g^n>$ is the subgroup generated by $g$. Let $\chi$ be a character on $G$ such that $\{\chi(x):x\in G\}$ is finite.
Then, can I conclude that $\chi$ is trivial on $G$?
So what I realized is this: take a sequence $g^{n_k}$ converging to $1$, then $\chi(g^{n_k})\to1$ but as range of $\chi$ is finite, we have $\chi(g^N)=1$ for some $N$ implying $\chi(g)^N=1$ implying $\chi(g)$ is an $N$-th root of unity.
Then I also observe that for any $x$, if we take $g^{n_j}\to x$ then by analogous argument as above, there exists $J_x$ with $\chi(x)=\chi(g)^{J_x}$.
So all points in range ($\chi$) lie on the $N$ vertices of a regular $N$ gon constructed inscribed in the unit circle.
No, not at all. For a trivial counterexample, let $G$ be a finite cyclic group.
More generally, an element $g\in G$ which generates a dense subgroup is equivalent to a homomorphism $\mathbb{Z}\to G$ with dense image. By Pontryagin duality, this is equivalent to an injective homomorphism $\hat{G}\to S^1$, where $\hat{G}$ is the character group of $G$. Your question then becomes: is every subgroup of $S^1$ torsion-free? The answer is obviously no.