In fact, uniform convergence on compact subsets of $\mathbb{Q}$ \mathbb{Q}\subset\mathbb{R}$ induces the usual topology on its group of (continuous) characters $\mathbb{R}\simeq\{t\mapsto\exp(ixt)\}_{x\in\mathbb{R}}$.
Namely, consider $K=\{0\}\cup\{1/n,n\geq 1\}$. If For $x$ x\in\mathbb{R}$, the corresponding character is real and uniformly $\epsilon$-close on $K$ to the trivial character iff $$|exp(ix/n)−1|<\epsilon\;\;\;\; (*)$$ for all integers $n\geq1$, then n\geq1$. Then for small enough $\epsilon$ ($<1/\sqrt{2}$ will do), \epsilon<1/\sqrt{2}$, $x$ must be small : $|x|<2\epsilon/\pi$ (hint: <2\epsilon/\pi$. Indeed, consider $k\in\mathbb{Z}$ such that $|x−k\pi|\leq\pi/2$ , and take $n=|k|$ n=|k|$; if $k\neq 0$ to we reach a contradiction in $(*)$. Hence $k=0$, and the claim follows easily)easily.
This implies that uniform convergence on compact subsets of $\mathbb{Q}$ (in fact the one compact subset ) $K$) induces the usual topology on $\mathbb{R}$.\mathbb{R}\simeq\mathrm{Hom}(\mathbb{Q},S^1)$.
