Compactness problem: certain parts of solution unclear? I have been stuck, even after having seen the solution, on a question at the end of one of my problem sheets; it goes like this:
Question
Let $T$ be a compact Hausdorff space  and let $f:T \rightarrow T$ be continuous. Show that there exists a non-empty subset $A \subset T$ such that $f(A) = A$
Solution
"Let $A_1 = T$, $A_n = f(A_{n-1})$ and $A = {\textstyle \bigcap_i^{\infty}} A_n$. Then $A_1 \subset A_0$ so by induction $A_{n} \subset A_{n-1}$"
Why is the 'then' part true? Does it follow from $f(T) \subset T$ (if that is even true)? If so, then why do we start with $A_0$ when we don't even know what that is? That doesn't make sense to me.
In any case, after this he shows quickly that $f(A) \subset A$ which I do understand. It is when he proves the other way around that I get lost again; he writes:
"Suppose $x \in A := {\textstyle \bigcap_i^{\infty}} A_n$. Then $f^{-1}(x)$ is a closed set"
Why does the first part imply the second? 
"So for every $n$, $A_n \cap f^{-1}(x) \neq \emptyset$"
Does this follow from the Hausdorff property? If so, I am confused; I thought that all that property stipulates is that any two distinct points are contained in disjoint open sets. 
Again, after this initial stage, the conclusion that $A \subset f(A)$ is clear to me. But if anyone could address these existing queries I would be extremely grateful. I think some of my questions have obvious answers but I can't think clearly enough right now to see that.
 A: Regarding the first part, that $A_1 \subseteq A_0$. This must be a typo. The point is they are defining a sequence of sets $A_1, A_2,\ldots$ where the first element is the whole space $T$ and since the function $f$ is from $T$ to $T$ we have by definition that the image of $T$ is a subset of $T$, i.e. $f(T) \subseteq T$ or $A_2 \subseteq A_1$ by definition of the sets $A_i$.
Regarding 

Suppose $x \in A := \bigcap_{n} A_n$. Then $f^{-1}(x)$ is a closed set.

Your assumption is that the space $T$ is Hausdorff. This in particular implies that every singleton $\{x\} \subseteq T$ is a closed set (compactness is not needed for this). And since $f$ is continuous the preimage of any closed set is closed. So in particular $f^{-1}(x)$ is closed.
Regarding

So for every $n$, $A_n \cap f^{-1}(x) \neq \emptyset.$

This follows from the way the sets $A_n$ are defined. To see that all these sets are inhabited you need to show there exists a $y \in A_n \cap f^{-1}(x)$. Or equivalently, there exists a $y \in A_n$, such that $f(y) = x$.
But by definition $x \in \bigcap_{n}A_n \subseteq A_{n+1} = f(A_n)$. So by definition of the set $A$ there exists such a $y$.
A: It looks like when they wrote $A_0$ that was a typo. It should have started with $A_1$. So don't worry about that. Additionally, note that for any sets $A$ and $B$, it's true that $f$ respects the subset relation. That is, $A \subset B$ implies $f(A) \subset f(B)$. This is easy to check. 
For your next question, note that the singleton $\{x\}$ is compact. Now, think about when compact implies closed...
And for the next part, you don't need the hausdorf property. Just think about the definition of $A_n$ and where $x$ lives. 
A: If $x\in A$, then we have $x\in f(A_n)$ for every $n$. Hence every $A_n$ contains some point mapped to $x$, so $f^{-1}(x)\cap A_n\ne\emptyset$. Note that $\{x\}$ is closed because for any point $y\ne x$, we can find a neighborhood of $y$ disjoint from (some neighborhood of) $x$. Thus $f^{-1}(x)$ is closed, and so is every $f^{-1}(x)\cap A_n$ since $A_n$ is compact and thus closed.
And I think they meant $A_0=T$. Either that or they should start with $A_2\subseteq A_1$.
