Suppose that a non-empty Hausdorff space $X$ is compact and there is a continuous map $f:X \to X$. Let $X_1=X$ and put $X_{n+1} = f(X_n)$ inductively for all $n \in \mathbb{N}$.
Prove that $A = \bigcap\limits_{n=1}^\infty X_n$ is non-empty.
Thoughts: Since $f$ is continuous, we have that $X_{n}$ is compact for all $n$. Furthermore, they are all closed because of $X$'s Hausdorffness.
In a normal metric space this would be easier, using the equivalence of compactness and sequential compactness, by constructing a sequence such that $a_n \in X_n$, and using the fact that it necessarily has a convergent subsequence with limit in the intersection.
Any hints would be really helpful.
Thanks!
Hint. Let $U_n := X \setminus X_n$. If we have $\bigcap_n X_n = \emptyset$, we would have $\bigcup_n U_n = X$ with open $U_n$. Now use compactness.