real analysis - $f(bigcap K_n)=bigcap f(K_n)$? Where $K_n$ are compact

$f:X\rightarrow Y$ is continous map between metric spaces, $K_n$ are non empty nested sequence of compact subsets of $X$, then we need to show the title above.

Please tell me which result I should apply here? regarding cont map and compact set I know that image of compact set is compact, attains bounds, uniformly continous etc. please help. well, we can start by taking $y\in \bigcap f(K_n)$ and then show that it is also in the left side?

Answer

Let $y \in \bigcap f(K_n)$. This means there is $x_n$ so that $f(x_n) = y$ and $x_n \in K_n$ for all $n$.

If $\{x_n\}$ is finite, we're done. If it's infinite, it has a limit point $x$ (why?). Use the continuity of $f$ to find $f(x)$ and conclude the proof.