This is from Forster's book "Lectures on Riemann surfaces"
Thorem 8.4
$X$ and $Y$ are Riemann surfaces. $A \subset X $ is a closed discrete subset. $ \pi' : Y' \to X - A $ is a proper unbranched holomorphic covering map.
Then $\pi '$ extends to a branched covering of X, i.e. there exists a Riemann surface $Y$, proper holomorphic map $\pi : Y \to X $ and a fiber preserving biholomorphic map $ \phi : Y - \pi ^{-1}(A) \to Y' $
Proof.
For every $a\in A$ choose a coordinate neighborhood $(U_{a}, z_{a})$ of $a$ on $X$ with the following properties : $z_a(a)=0$ and $z_a(U_a)$ is a unit disk in $\mathbb{C}$ and $U_a \cap U-{a'} = \varnothing $ if $a ≠a'$. Let $U_a ^* = U_a - \{a\}$.
Since $\pi : Y' \to X' $is proper, $ \pi ' ^{-1}(U_a ^*) $ consists of a finite number of connected components $V_{a, \nu} ^*$ $ \nu = 1, ... , n(a) $ For every $\nu$, $ \pi ' |V_{a, \nu} ^* \to U_a ^*$ is a proper unbranched holomorphic covering. ...
Question : I don't understand why $ \pi ' ^{-1}(U_a ^*) $ should have finitely many components. I know that $\pi '$ has finitely many sheets, say $m$ sheets, since it is proper. Thus every $x \in U_a ^* $ has an evenly covered connected neighborhood of which inverse image is disjoint union of $m$ connected components. Do these local inverse images piece together nicely to make $ \pi ' ^{-1}(U_a ^*) $ consist of finite components?
Also I cannot see why for every $\nu$, $ \pi ' |V_{a, \nu} ^* \to U_a ^*$ is a proper unbranched holomorphic covering.
I would appreciate any hint or other references.