 Solution

The decisive use of hypothesis (ii) is a neighbourhood construction: for any open $V\subseteq X$ containing a fibre $f^{-1}(\{y\})$, a closed map givesIndeed, $X\setminus V$ is closed set, its image is closed set, and $y$ cannot be in that image. A point of $f^{-1}(W)$ cannot lie outside $V$. Applying this to the finite union covering each compact fibre supplies a whole neighbourhood of that fibre's value using the same finitely many cover elements. Compactness of $K$ selects finitely many such neighbourhoods and hence a finite subcover of $f^{-1}(K)$, proving the desired compact-preimage theorem for closed maps.

The closed map assumption cannot simply be omitted. Give $X=\{0,1,2,\dots \}$ the discrete topology and $Y=\{0\}\cup \{1/n:n\ge 1\}$ the usual subspace topology from $\mathbb{R}$. The bijection $f(0)=0$, $f(n)=1/n$ is continuous, and all its fibres are compact singletons. But $Y$ is compact whereas $f^{-1}(Y)=X$ is an infinite discrete space, which is not compact. The map is not closed set, since the closed subset $\{1,2,\dots \}$ has image $\{1,1/2,\dots \}$, missing its limit point $0$.