$f\left( \bigcap_{n=1}^{\infty}A_{n}\right) = \bigcap_{n=1}^{\infty}f(A_{n})$ for a continuous function on a compact space

Hint Let $y \in \bigcap_{n=1}^{\infty}f(A_{n})$. Then, there exists some $x_n \in A_n$ such that $f(x_n)=y$.

Now, by the compactness of $X$, $x_n$ has a convergent subsequence $x_{k_n}\to x$. By the continuity of $f$ you have $f(x)=y$.

Show that $x \in \bigcap_n A_n$.