Is there a profinite group $G$ which is not its own profinite completion?

Surely not, I thought. But upon looking into it, I found that there is a special name given to a $G$ which *is* its own profinite completion, namely "strongly complete". And a recent (2003) hard theorem (which according to Wikipedia uses the classification of finite simple groups) due to Nikolov and Segal asserts that, if $G$ is finitely generated (as a topological group), then it is "strongly complete".

So the $G$ I'm looking for cannot be topologically finitely generated. An equivalent question to the above is:

Is there a profinite group $G$ which admits a non-open subgroup of finite index?

Now here's my problem; the only exposure to profinite groups I've had has been in the context of number theory, absolute Galois groups, local fields, etc. In particular, the only non-topologically-finitely-generated profinite group I'm aware of is the absolute Galois group of a number field, say $\mathbb{Q}$. But I reckon the Krull topology demands that the finite index subgroups of $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ be open.

Maybe there is a more 'exotic' example of such a $G$...