I am self studying Profinite Groups by Ribes and Zalesskii and I am stuck reading a proof.
Let $G$ be a profinite group and $H \unlhd G$ a closed normal subgroup. If $H$ is finite then there exists $U \unlhd G$ open normal subgroup such that $U \cap H = 1$.
(This is the first passage in the proof that every epimorphism between profinite groups admits a section, page 29 in my edition)
As this is presented as a fact I think I am missing something easy, however I can't seem to prove it. I tried to reason in the case where all the groups are finite with the discrete topology, but then the assertion seems false: if $G$ is a finite $p$-group and $H = Z(G)$ then for each nontrivial normal subgroup $N \unlhd G$ we have $H \cap N \neq 1$.
What am I missing here?