The question is:
Suppose $f\colon X\to Y$ is continuous on a compact metric space $X$, $Y$ is a metric space and $C\subset Y$ is closed. Show that for any open neighborhood $U$ of $f^{-1}(C)$ in $X$, there exists an open neighborhood $V$ of $C$ in $Y$ such that $f^{-1}(V)⊂U$.
I have tried to argue that $f^{-1}(C)$ is closed (and hence compact) by continuity of $f$ and compactness of $X$, then $C=f(f^{-1}(C))$ is also compact by continuity of $f$ again.
But the compactness of the two sets seems don't help me very much. Can anyone help me? Thx!