You do need more than the fact that the map is open and has finite fibres and compact range.
For $n\in\Bbb Z^+$ let $$X_n=\left\{\left\langle\frac1n,k\right\rangle\in\Bbb R\times\Bbb Z:0\le k\le n\right\}\;,$$ and let $$X=\{\langle0,0\rangle\}\cup\bigcup_{n\in\Bbb Z^+}X_n\;,$$ topologized as a subset of $\Bbb R^2$ with the usual topology. Let $f:X\to\Bbb R:\langle x,y\rangle\mapsto x$ be the projection to the first coordinate, and let
$$Y=f[X]=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}\;;$$
clearly $Y$ is compact, and $f$ has finite fibres. I claim that $f$ is open.
To see this, let $U\subseteq X$ be open. If $\langle0,0\rangle\notin U$, then $f[U]\subseteq Y\setminus\{0\}$, so $f[U]$ is open in $Y$. And if $\langle0,0\rangle\in U$, there is an $m\in\Bbb Z^+$ such that $\left\langle\frac1n,0\right\rangle\in U$ for each $n\ge m$, in which case $f[U]$ contains a nbhd of $0$ in $Y$ and must again be open in $Y$.
Finally, $X$ is clearly not compact, as $\left\{\left\langle\frac1n,n\right\rangle:n\in\Bbb Z^+\right\}$ is a closed set in $X$ with no limit point.
Notice that this $f$ is not a covering map: the point $0$ in $Y$ has no evenly covered open nbhd. You really will need to use the fact that you’re dealing with a covering map.
Added: A map $f:X\to Y$ is perfect if it is continuous and closed and has compact fibres. If you prove that your covering map is closed, you’ll have shown that it is perfect. Now use the following lemma.
Lemma. If $f:X\to Y$ is perfect, and $Y$ is compact, then $X$ is compact.
Start of Proof. Let $\mathscr{U}$ be an open cover of $X$. For each $y\in Y$ let $X_y=f^{-1}\big[\{y\}\big]$; by hypothesis each $X_y$ is compact, so there is a finite subset $\mathscr{U}_y$ of $\mathscr{U}$ that covers $X_y$. Let $V_y=\bigcup\mathscr{U}_y$; clearly $X_y\subseteq V_y$. Use the fact that $f$ is closed to show that there is an open $W_y\subseteq Y$ such that $X_y\subseteq f^{-1}[W_y]\subseteq V_y$. $\mathscr{W}=\{W_y:y\in Y\}$ is then an open cover of $Y$, so it has a finite subcover $\mathscr{W}_0$. Use $\{f^{-1}[W]:W\in\mathscr{W}_0\}$ and the appropriate collections $\mathscr{U}_y$ to find a finite subcover of $\mathscr{U}$.
Consider for example the $2$ point space $\{x,y\}$ with the discrete topology. The space is covered by $\{x,y_1,y_2\}$ (discrete topology) where $p(x)=x$ and $p(y_i)=y$.
Now $x\in U_x=\{x\}$ and $y\in V_y=\{y\}$. Now $p^{-1}(U_x)=\{x\}$ and $p^{-1}(U_y)=\{y_1\}\cup\{y_2\}$.
Best Answer
$\Rightarrow$ Each fiber is compact (by properness) and discrete (from definition of covering space) hence is finite.
$\Leftarrow$ You have to prove that for $K\subset X$ the inverse image $q^{-1}(K)$ is compact.
Since $\operatorname {res} q:q^{-1}(K) \to K$ is a finite covering space in its own right apply my answer to cocomi.