general topology - Let $K_n$ is a nonempty compact subset of $X$ and that $K_{n+1} subset K_n$ for each $n in Bbb N$. Show that $f(K) = bigcap _n f(K_n)$.

Let $f: X \to Y$ be a continuous map between metric spaces. Assume that $K_n$ is a nonempty compact subset of $X$ and that $K_{n+1} \subset K_n$ for each $n \in \Bbb N$. Let $K := \bigcap _n K_n$. Show that $f(K) = \bigcap _n f(K_n)$.

I have done in the following way:

$\Rightarrow$
If $x \in K$, then $f(x) \in f(K_n)$ for all $n$ and so $f(K) \subset \bigcap f(K_n)$.

$\Leftarrow$

Let $y \in f(K_n)$ for all $n$. The set $f^{-1}(y) \bigcap K_n$ is a nonempty closes subset of $K_n$ and hence compact. Also they are decreasing and hence by Cantor's Intersection Theorem, $\bigcap (f^{-1}(y) \bigcap K_n) \neq \phi$. This means that there exist $x \in f^{-1}(y) \bigcap K_n$ for all $n$.

Is the solution correct. Does it have a better solution??