Remark

A filter which is coarser than a filter with a countable base need not possess a countable base; for example, if $X$ is an uncountable infinite set, then the filter consisting of the complements of finite subsets of $X$ has no countable base (otherwise the set of finite subsets of $X$ would be countable, contrary to assumption); nevertheless this filter is coarser than every elementary filter associated with an infinite sequence of distinct elements of $X$.