Restriction of Continuous Function on Compact Hausdorff Space I'm trying to show that given a compact Hausdorff space $X$ and given a continuous function $f:X\to X$, there exists a compact subset $K\subseteq X$ such that $f_{\big|K}:K \rightarrow X$ maps $K$ onto $K$.
I'm not sure how to approach this.
 A: We could argue that $K=\emptyset$ solves the problem.
This part is wrong, or inconclusive at best: I'll rectify it in the following paragraph. For a non-empty subset, let's preliminarily consider a hypothetical non-empty compact subset $Q$ such that $f[Q]\subseteq Q$. Then, you can consider the family of compact sets $Q_0=Q$ and $Q_{n+1}=f[Q_n]$: in other words, $Q_n=f^n[Q]$. By the hypothesis that $f[Q]\subseteq Q$, we have that $Q_{n+1}\subseteq Q_n$ for all $n$, which implies that this is a decreasing sequence of closed ($X$ is Hausdorff) non-empty compact set. Therefore its intersection $K=\bigcap_{n\in\Bbb N} Q_n$ is non-empty. Moreover, $f[K]=K$ (not necessarily, that I know of: as it's been pointed out, only $f[K]\subseteq K$ is immediate).

Rectification of part 1:
As it has been pointed out, the procedure used to devise a non-empty subset $K$ such that $f[K]=K$ starting from some non-empty compact $Q$ such that $f[Q]\subseteq Q$ needs to be amended.
Let $\kappa$ be an inital ordinal strictly larger than $\lvert X\rvert$ and define by transfinite induction the generalised sequence $Q_\bullet: \kappa+1\to \mathcal \{\text{compact subsets of }X\}$ $$\begin{cases}Q_0=Q\\ Q_{\beta+1}=f[Q_\beta]\\ Q_{\beta}=\bigcap_{\gamma<\beta} Q_\gamma&\text{if }\beta\text{ is a limit ordinal}\end{cases}$$
Notice that the hypothesis that $X$ is Hausdorff is needed for the sequence $Q_\bullet$ to be well-defined, i.e. to guarantee that its range stays in the family of compact subsets of $X$.
Now, it is clear that $Q_\bullet$ is weakly decreasing. It cannot be strictly decreasing, because otherwise $\lvert X\setminus Q_\kappa\rvert\ge \kappa>\lvert X\rvert$. Therefore, there is some $\beta$ such that $Q_\beta=Q_{\beta+1}$. $K=Q_\beta$ satisfies $f[K]=K$ by definition, so let's pick the one corresponding to the least such ordinal $\beta$.
We just need to prove that $Q_\beta\ne \emptyset$. If $\beta=\gamma+1$, for some ordinal, then we are ok, because $f[Q_\gamma]=\emptyset$ implies $Q_\gamma=\emptyset$, against minimality of $\beta$. If $\beta$ is a limit ordinal, then $\bigcap_{\gamma<\beta} Q_\gamma=\emptyset$ implies that $\{X\setminus Q_\gamma\}_{\gamma<\beta}$ is an open (recall that $X$ is Hausdorff) cover of $X$. Therefore there is a finite subcover $\{X\setminus Q_{\gamma_1},\cdots, X\setminus Q_{\gamma_t}\}$, say, with $\gamma_1<\cdots<\gamma_t<\beta$. But since $X\setminus Q_{\gamma_1}\subseteq\cdots\subseteq X\setminus Q_{\gamma_t}$, we have that $Q_{\gamma_t}=\emptyset$ and $f\left[Q_{\gamma_t}\right]=\emptyset$, against minimality of $\beta$.

This procedure may be used on the whole space $X$. Another way to find a starting compact is to select a non-empty subset $U$ and consider $U^f:=\bigcup_{n\in\Bbb N} f^n[U]$ and  $Q=\overline{U^f}$. Notice that $f\left[U^f\right]\subseteq U^f$, and therefore $f[Q]\subseteq \overline{U^f}=Q$.

Remark: The transfinite induction I've made uses choice in its cardinality argument. However, that passage may be avoided by simply having $\kappa$ be an ordinal which does not inject into the family of compact sets of $X$ (for instance, the Hartogs number of $\mathcal P(X)$). Then $Q_\bullet$ cannot be strictly decreasing, because it cannot be injective.
