Example taken from Ribes and Zalesskii's book "Profinite groups". Take an infinite set $I$ and a finite group $T$. You can let $G$ be the profinite group $\prod_I T$. Denote its elements by $(g_i)_{i\in I}$. Let $\mathcal F$ be an ultrafilter which contains the filter of all cofinite subsets of $I$. If you denote $H$ to be the subgroup of elements with $\lbrace i\in I\ |\ h_i=1\rbrace\in g_i=1\rbrace\in \mathcal F$, it is clear that $H$ is proper normal and that it is not open because it is dense and has finite index $|T|$.
To show that $H$ has index $|T|$ in $G$ consider all elements $a_t=(t,t,\dots)$ for $t\in T$. For any $g\in G$, consider $I_t=\lbrace i\in I \ | \ g_i=t\rbrace$. Since we have $\bigcup_{t\in T} I_t=I$ then $I_t\in \mathcal F$ for some $t$, and therefore $ga_t^{-1}\in H$.
Example taken from Ribes and Zalesskii's book "Profinite groups". Take an infinite set $I$ and a finite group $T$. You can let $G$ be the profinite group $\prod_I T$. Denote its elements by $(g_i)_{i\in I}$. Let $\mathcal F$ be an ultrafilter which contains the filter of all cofinite subsets of $I$. If you denote $H$ to be the subgroup of elements with $\lbrace i\in I\ |\ h_i=1\rbrace\in \mathcal F$, it is clear that $H$ is proper normal and that it is not open because it is dense and has finite index $|T|$.