Is it provable, in ZF (without Choice), that every filter can be extended to one of cardinality continuum?
The extended filter is not requested to be an ultrafilter.
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If we talk about filters on $\omega$, then the answer is an easy yes.
If $\cal F$ is a filter on $\omega$ such that there is at least one $A\in\cal F$ which is co-infinite, then for every $B\subseteq\omega\setminus A$ we have that $A\cup B\in\cal F$.
But every filter can be extended (perhaps trivially) to include at least one co-infinite set. Otherwise all the co-infinite sets are in the co-ideal, which means that the filter is empty.
If we are talking about general filters, then the answer is negative even with choice, since taking any uniform filter on a large enough set gives a counterexample. Or filters on finite sets.
If we are talking about general filters, but you mean "filters on infinite sets" and "at least continuum", then the answer is consistently negative in the absence of choice. If $A$ is a set with a Dedekind-finite power set, then any filter on $A$ cannot have cardinality greater or equal to the continuum. Such examples include amorphous sets.