Suppose $\ f \colon X \to X $ is a continuous function from a compact,Hausdorff space to itself. Prove that there exists a subspace $A$ such that $f(A) =A$.
I came up with an answer based on nets (see the answer below). But i am not very much familiar with nets. So please point out,if there is any error in argument.
Let $Y=f(X)$ then $X= \mathop{\cup}_{y \in Y} f^{-1}(y)$ where the union is over disjoint sets. Since each of the pullback is a closed set. Being in a normal space, $\forall \ f^{-1}(y), \ \exists \ U_{y}$ open in $X$ such that $f^{-1}(y) \subset U_{y} \subset \overline{U_{y}} \subset X$ .
Hence $X = \bigcup_{y \in Y} U_{y} $. This open cover has a finite subcover. $X=U_{1} \cup U_{2} \cup \ldots \cup U_{n} $. Let $\Sigma = \lbrace A \ \ \vert A \subset X \rbrace$, suppose $\nexists$ any $A \in \Sigma$ st $f(A) =A$ then $\forall A \in \Sigma , \ \exists \ f (x)_{A} $ st $f (x)_{A} \notin A$.
define partial order on $\Sigma$ by set inclusion so that it becomes a directed set.
Let $\ell \colon \Sigma \to X$ be a net such that $\ell(A) = f(x)_{A} $. For simplicity let us represent it the net by $(f(x)_{A})$.
Since $X$ is compact,every net has a cluster point,call the cluster point of the above net as $p$. Since $p \in X=U_{1} \cup \ldots \cup U_{n} $, we may assume $p \in U_{1} $.
Let $U_{1}$ be the neighbourhood of $p$ and $A= U_{1}$ then $\exists \ B \in \Sigma $ such that $ A \subset B $ and $f(x)_{B} \in U_{1}$. But $f(x)_{B} \notin B \Rightarrow f(x)_{B} \notin A=U_{1}$.
This is a contradiction. $\square$