Definition: Let $G$ be a topological group. We call $G$ a finite quotient of $G$ if there exists a normal subgroup $N$ of $G$ such that $H = G/N$ and $H$ is finite.
Let $G'$ be a dense subgroup of $G$.
Question: Do $G$ and $G'$ have the same finite quotients?
Observations and remarks: In some notes I found some examples where this is true, e.g. for $G' = \mathbb{Z} \subset G = \hat{\mathbb{Z}} = \varprojlim{\mathbb{Z}/n\mathbb{Z}}$ and $G' = W_K \subset G = G_K = \operatorname{Gal}(\bar{K}/K)$ where $K$ denotes a local field, $G_K$ the absolute Galois group and $W_K$ the Weil group. But there the authors have never given any proofs, so I don't know if this holds in general. My knowledge on topological groups is also not worth mentioning as I have no clue what kind of arguments would be helpful. It was just something that made me suspicious.
Could you please help me answering my question? Thanks a lot!
No. For example, $S^1$ has a dense subgroup isomorphic to $\mathbb{Z}$ but $S^1$ has no nontrivial finite quotients.
What is special about the case of $\mathbb{Z}$ and the profinite integers is that the latter is the profinite completion of the former.