Here is a partial answer: if $|X|^{\aleph_0}=|X|$ then the answer is yes. First take an injective function $F:[X\times X]^\omega\to X$ and then take some function $G:[X\times X]^\omega\to X$ such that for all $A\in[X\times X]^\omega$ the point $\langle F(A),G(A)\rangle$ is not in $A$. Translate this via a bijection between $X$ and $X\times X$ to a function $H:[X]^\omega\to X$ such that $H(A)\notin A$ for all $A$.

Then $f:A\mapsto X\setminus\{H(A)\}$ is as required.

Here is a partial answer: if $|X|^{\aleph_0}=|X|$ then the answer is yes. First take an injective function $F:[X\times X]^\omega\to X$ and then take some function $G:[X\times X]^\omega\to X$ such that for all $A\in[X\times X]^\omega$ the point $\langle F(A),G(A)\rangle$ is not in $A$. Translate this via a bijection between $X$ and $X\times X$ to a function $H:[X]^\omega\to X$ such that $H(A)\notin A$ for all $A$.

Then $f:A\mapsto X\setminus\{H(A)\}$ is as required.

Note that this uses nothing but the $T_1$-property and works even for the co-finite topology.