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 $\mathscr{U}$.
Firstly, as a notation issue $\mathcal{U}_X$ is a cover of $X$ and likewise $\mathcal{U}_Y$ is one for $Y$. We are looking for a subcover of $\mathcal{U}$ which are open sets in $X \times Y$, while your $\mathcal{U}_X$ are open sets in $X$, not of $\mathcal{U}$.
You probably mean the sets in $\mathcal{U}$ they are projections of, so if we have $\mathcal{A}' = \{\pi_1[U_1], \ldots, \pi_1[U_n], \ldots, \}$ as a countable subcover of $\mathcal{A}$ you define $\mathcal{U}_X$ as $\{U \in \mathcal{U}: \pi_1[U] \in \mathcal{A}'\}$, so at least all the $U_n$ etc. Similarly for $\mathcal{U}_Y$ and a finite subcover $\mathcal{B}'$ of $\mathcal{B}$, of course.
How would a "proof" that this is in fact a subcover go? Suppose $(x,y) \in X$. Then $x$ is covered by some $\pi_1[U(x)] \in \mathcal{A}'$, and also by some $\pi_1[U(y)] \in \mathcal{B}'$. Nothing garantuees you that $(x,y)$ is in either $U(x)$ or $U(y)$ as the first only has a condition only on $x$ and the second only on $y$, not on them both combined.
You might think, define my subcover to be all products $\{U \times V: U \in \mathcal{A}', B \in \mathcal{B}'\}$. This is not necessarily a subcover either: e.g. suppose we were covering with open circles in the plane, then the products of projections give open squares (so a totally different open cover).
The final idea you could have (also doesn't work): start with a cover $\{U_i \times V_i: i \in I\}$ of basic open sets (these suffice for showing Lindelöfness), find countably many $J \subset I$ such that $\{U_i: i \in J\}$ cover $X$ and finitely many $J'$ such that $\{V_j: j \in J'\}$ covers $Y$.
Then form all $\{U_i \times V_j: i \in I', j \in J'\}$. These now do form a countable cover of $X \times Y$, but not necesarilly a subcover as the index sets $I',J'$ could be totally disjoint (we get them independently of each other), so we are making new sets here that were not in the original cover, which is what the point was.
It's good to think about proofs like this; they don't work, and you see why you need another idea (here the tube lemma).
Best Answer
I think there might be problems in your last line of reasoning.
First, let me settle some notation. If $O_1,\dotsc, O_n$ is your original finite cover, then let $U_i$ be the compactly included refinement. We have $K\subset \bigcup_i U_i\subset C\subset \bigcup_i O_i$ where $C$ is the union of the closures of the $U_i$. Also, we put $V_i =C\setminus U_i$, which are each compact.
I think your desired argument goes something like this: if $z\in K_\delta$ we have $d(z,x)<\delta$ for some $x\in K$. On the other hand, $d(x,V_i)\geq\delta$ for some $i$ (since otherwise $f(x)<\delta$, contradicting that $\delta$ was the minimum). Now you want to say $z\in U_i$, since otherwise $z\in V_i$ and we get $d(x,V_i)\leq d(x,z)<\delta$, a contradiction. This only works though if we assume $z\in C$ to begin with. The most we can say is that $z\in (C\setminus U_i)^c = C^c\cup U_i$.
Here's one possible fix: define the $V_i$ to be $U_i^c$ instead of $C\setminus U_i$. After all, we don't need $V_i$ to be compact for $x\mapsto d(x,V_i)$ to be continuous. And we still know $f(x)>0$ for $x\in K$ since otherwise $x\in V_i$ for each $i$, which would in turn mean $x\notin \bigcup U_i\supset K$.
Honestly though, this result can be proven with much less machinery. Let ${\cal O}= \{O_\alpha: \alpha\in A\}$ be an open cover of $K$. For each $x\in K$, there is some element of this collection $O_{\alpha_x} = O_x$ which contains $x$. Also, by openness, there is some $\delta_x>0$ such that $B(x,2\delta_x)\subset O_x$.
Now it's not hard to see that the collection $\{B(x,\delta_x): x\in K\}$ is an open cover of $K$, and hence there are finitely many points $x_1,\dotsc, x_n$ such that $$K\subset B(x_1,\delta_1)\cup\dotsm\cup B(x_n,\delta_n).$$ Yet now we've given ourselves some wiggle room. Denoting $O_k = O_{x_k}$, we actually have $$K\subset B(x_1,\delta_1)\cup\dotsm\cup B(x_n,\delta_n)\subset B(x_1,2\delta_1)\cup\dotsm\cup B(x_n,2\delta_n)\subset O_1\cup\dotsm\cup O_n.$$ From here, put $\delta = \min (\delta_1,\dotsc, \delta_n)$. You can easily see by triangle inequality that each point of $K_\delta$ belongs to one of the $B(x_k,2\delta_k)$ balls, and in particular is covered by $O_k$.