I am currently reading through a proof of the existence of a right haar measure for locally profinite groups.
We have a locally profinite group $G$ with the assumption that for all compact open subgroups $K$, $G/K$ is countable.
In the proof, the authors take a decreasing sequence of compact open subgroups $\{K_n\}_{n \geq 1}$ with $\cap_{n \geq 1} K_n =1$.
Why can they do this? Maybe I'm missing something obvious.
If the identity admits a countable local basis of compact open subgroups then this is true.
It's indeed false. A locally profinite group admits such a sequence if and only if it's metrizable (i.e., has a second countable compact open subgroup).
Indeed, if it admits such a sequence, then the map from $K_0$ into the metrizable product $\prod_n K_0/K_n$ is injective continuous, hence $K_0$ is metrizable, so $G$, which is homeomorphic to $K_0\times (G/K_0)$, is metrizable.
Conversely, if $G$ is metrizable, so is $K_0$, which being compact metrizable admits only countably many clopen subsets, whence the result.
For instance, for every nontrivial finite group $F$ and uncountable set $X$, the uncountable power $F^X$ is a counterexample. So the authors of your book just forgot to assume $G$ metrizable (or they did and you didn't notice it). Whether their proof can be adapted otherwise, I don't know.