Partition of sets in topological spaces that preserves limit point for each member

Recall that for a sequence $$\{x_{n}\}$$ in a metric space, if $$\{x_{n}\} \rightarrow x$$ while $$x_{n} \neq x \ \ \forall n$$, then $$\{x_{2k}\} \rightarrow x$$ and $$\{x_{2k+1}\} \rightarrow x$$. As a corollary, for arbitrary limit point $$a$$ of a countable set $$A$$, there exists an n-partition $$\{A_{1}, ... ,A_{n} \}$$of $$A$$ s.t. $$t \in \overline{A_{i}},\forall i \in {1,...,n}$$.

I wish to generalize this to arbitrary compact Hausdorff spaces and arbitrary sets (without assuming countability). More precisely, I wish to prove:

Given a compact Hausdorff space $$X$$, a subset $$A$$ of $$X$$, and a point $$x \in \overline{A}-A$$. There exists a partition $$A = A_{1} \sqcup A_{2}$$ such that $$x \in \overline{A_{1}} \cap \overline{A_{2}}$$.

Note that direct application of sequential methods fails since $$X$$ might not be first-countable.

This is not true. For instance, let $$X=\beta\mathbb{N}$$, the Stone-Cech compactification of $$\mathbb{N}$$ with the discrete topology, let $$A=\mathbb{N}$$, and let $$x$$ be any point of $$\beta\mathbb{N}\setminus\mathbb{N}$$. Suppose $$A=A_1\sqcup A_2$$ is any partition. Then there is a continuous function $$f:\mathbb{N}\to[0,1]$$ which is $$0$$ on $$A_1$$ and $$1$$ on $$A_2$$, and this function extends continuously to $$X$$. If $$x$$ were in the closure of both $$A_1$$ and $$A_2$$, then the continuous extension would have to take both the value $$0$$ and the value $$1$$ at $$x$$.
In general, if $$x\in\overline{A}$$, the set of neighborhoods of $$x$$ intersected with $$A$$ forms a filter $$F$$ on $$A$$. If $$B\subseteq A$$, then $$x\in\overline{B}$$ iff the complement of $$B$$ is not in $$F$$. So, your question is whether there exists a set $$A_1\subseteq A$$ such that neither $$A_1$$ nor its complement is in $$F$$. Such an $$A_1$$ exists iff $$F$$ is not an ultrafilter.