It's a known result that in an uncountable set $X$ with the topology of countable complementaries (i.e. $U$ is open if $X\backslash U$ is countable), a subset $A\subset X$ is compact if and only if $A$ is finite.
I thought of a proof of this fact. If $A$ is infinite, then let $B\subset A$ a countable set, and for each $x \in B$ let be $U_x = (X\backslash B) \cup \{x\}$, which is open because $B$ is countable. Then $A\subset\cup _{x\in B} U_x$ and it does not admit a finite subcovering. Thus, $A$ is not compact.
The thing is that for proving these I used the axiom of choice (only in $\mathtt{ZF}$ it's not sure that every set will have a countable subset).
My question is, ¿is this result true just on $\mathtt{ZF}$ (without $\mathtt{AC}$)?
It is not true in ZF alone.
It is consistent with ZF that there exist infinite, Dedekind-finite sets - these are sets which admit no non-surjective self-injection but are nonetheless infinite. It's a good exercise to show that Dedekind-finiteness is equivalent to not containing a countable infinite subset; so on such a set, the co-countable topology is just the cofinite topology. And the cofinite topology on any set is compact.
In fact, the statement "The cocountable topology on an infinite set is never compact" is exactly equivalent to the statement "There are no infinite Dedekind-finite sets."