Is every first countable profinite group actually second countable?
Although this question has been answered already by Masked Avenger, it seems like a good idea for me to explain why every firstcountable Hausdorff group is metrizable. If $G$ is a Hausdorff topological group and $U$ is an open neighborhood of the identity $e$, then let $R_{U}$ be the binary relation on $G$ where $(x,y)\in R_{U}$ iff $xy^{1}\in U$. Then $G$ becomes a Hausdorff uniform space with a uniformity generated by the relations $R_{U}$ where $U$ ranges over all open neighborhoods of the identity $e$. Furthermore, it is well known that a uniform space $(X,\mathcal{U})$ is induced my a metric if and only if $\mathcal{U}$ is generated by countably many relations (a proof of this fact is given in this answer). Therefore, if a $G$ is a first countable Hausdorff topological group, then the uniformity generated by the relations of the form $R_{U}$ is metrizable. Therefore, we conclude that every Hausdorff first countable topological group is metrizable. 


J. Wilson, Profinite groups, 4.1.3: a profinite group is metrizable iff it is a countable inverse limit of finite groups. Any firstcountable, Hausdorff group is metrizable. So, yes, every firstcountable profinite group is secondcountable, this is why they are usually called "separable". 

