I know that an infinite sigma algebra has to be uncountable and I was wondering if a similar statement holds for filters. I think that I can say that given an infinite set $X$, the principal filter generated by $x$ must be uncountable.
I think that given a finite set $F\subseteq X$ where $\vert X\vert \geq \aleph_0$, the principal filter $\mathcal{F}_F$ generated by $F$ must be uncountable, because we can construct an injection from $2^{X\setminus F}$ to $\mathcal{F}_F$.
I also think that going by this logic, if $\mathcal{F}$ is a set filter over $X$ such that $A\in \mathcal{F}$ and $\vert X\setminus A\vert\geq \kappa$, then $\vert \mathcal{F}\vert \geq 2^\kappa$.
I have the following questions:
$(i)$ Do these arguments seem right?
$(ii)$ Can we also conclude that if $\mathcal{F}$ is an infinite filter, then is $\mathcal{F}$ necessarily uncountable?
Your first argument is correct. But in order to prove that every infinite filter is uncountable, using the first argument, you'd have to prove that in every infinite filter there is a co-infinite set.
Unfortunately, for your argument, this isn't the case. Easily, $\{A\subseteq X\mid X\setminus A\text{ is finite}\}$ is a filter. And indeed, if $X$ is countable, it is a countable filter.