Topological contraction on compact spaces This is a follow up question. You can see the original here. I have the following problem.

Let $X$ be a compact Hausdorff space and let $f:X\to X$ be continuous.
  Show that there exists a non-empty set $A\subset X$ such that
  $f(A)=A$.

With the hint to define $A=\cap_{n\ge 0} A_n$ with $A_{n+1}=f(A_n)$ and $A_0=X$.
I have most of the solution by now, but there is one last thing that I cannot show and it drives me crazy.
It is obvious that $A_n$ is compact and by induction we have $A_{n+1}\subset A_n$, so $A$ is also compact and non empty.
The it needs to be shown that $f(A)=A$. One inclusion is easy since $x\in A$ implies that $x\in A_n$ for all $n$ so $f(x)\in A_{n+1}$ for all $n$, thus $f(x)\in A$. This shows the inclusion $f(A)\subset A$.
For the other inclusion, $A\subset f(A)$, I need to show that for all $x\in A$ implies the existence of $y\in A$ such that $f(y)=x$.
However $x\in A$ implies that $x\in A_{n+1}$ for all $n$, this implies that for any $n$ exists $y_n\in A_n$ such that $f(y_n)=x$. So I get the existence of a sequence of points with this property, but the space is not metric, so compactness does not imply sequential compactness, so I cannot get a converging subsequence.
My idea was to assume that the sequence does not converge and try to construct a countable cover of $X$ that does not admit a finite subcover, but I cannot see how to do that. I keep thinking in terms of distance or measure. I tried to work with the sets $X\backslash A_n$ but I couldn't define anything useful.
 A: You may not have sequential compactness, but compactness does imply that every infinite subset has a limit point. In order to utilize this, you need to first mention the case where you don't have an infinite set of points:


*

*If some $y \in X$ is equal to $y_n$ for infinitely many $n$, then $y \in A$ and $f(y) = x$ and you're done.


Otherwise, the set $\{y_0, y_1, y_2, \cdots\}$ is an infinite subset of $X$, hence has a limit point $y \in X$. We want to claim two things: $y \in A$, and $f(y)=x$.
If $y \notin A$, then there exists $n$ so that $y \notin A_n$, which means $X \setminus A_n$ is an open neighborhood of $y$ which excludes all points except possibly $\{y_0, \cdots, y_{n-1}\}$. Using the fact that $X$ is Hausdorff, you can extend this to contradict the assumption that $y$ is a limit point of $\{y_0, y_1, y_2, \cdots\}$.
If $f(y) \neq x$, then again use the fact that $X$ is Hausdorff to get a contradiction. In fact, if $U$ and $V$ are disjoint neighborhoods of $f(y)$ and $x$, then can you show that $y \in f^{-1}(U)$ and $y_0, y_1, y_2, \cdots \in f^{-1}(V)$? Why is this a contradiction?
