Showing whether a certain subset is evenly covered by a map I am stuck in problem 53.4 from Munkres: Let $ q : X \to Y $ and $ r : Y \to Z $ be covering maps; let $ p = r \circ q $. Show that if $ r^{-1}(z) $ is finite for each $ z \in Z $, then $ p $ is a covering map.
This is my work so far:

First, $ p $ is both continuous and surjective, because the
  composition of continuous/surjective maps is again
  continuous/surjective (note that both $ r $ and $ q $ are continuous
  and surjective, since they are covering maps).     
Let $ z \in Z $.
  Since $ r $ is a covering map, there exists an open neighborhood $ U $
  of $ z $ that is evenly covered by $ r $. For all $ y \in r^{-1}(z) $,
  there exists an open neighborhood $ V_{y} $ of $ y $ that is evenly
  covered by $ q $, since $ q $ is a covering map. We now define 
  \begin{align*} U' := \bigcap_{y \in r^{-1}(z)} r(V_{y}) \end{align*}
  We claim that $ U' $ is both open and evenly covered by $ p $, making
  $ p $ a covering map (since for all elements of $ Z $, a corresponding
  $ U' $ can be constructed).   
$ U' $ is open in $ Z $ because it is a
  finite intersection of finitely many open subsets $ r(V_{y}) $, as $ r^{-1}(z) $ is finite. Each $ r(V_{y}) $ is open in $ Z $ because the
  covering map $ r $ is open. Since each $ V_{y} $ is
  open, it follows that $ r(V_{y}) $ is open.      We now show that $ U' $
  is evenly covered my $ p $, that is, $ p^{-1}(U') $ is a union of
  disjoint open sets in $ X $. 

I am stuck in showing whether $p^{-1}(U')$ can be expressed as a disjoint union of open sets in $X$, each of which is homeomorphic to $U'$. Is it true that my $U'$ is evenly covered by $p$?

Continuity and surjectivity are clear, as you state. The work is in the evenly covered part of the definition:
Take a fixed $z \in Z$. This has the points $y_1, \ldots y_n$ (for some finite $n$ by assumption) as preimages in $Y$ and an evenly covered open neighbourhood $U_z$ such that for every $i=1,\ldots,n$ we have open $V_i$ pairwise disjoint and containing $y_i$ such that $r|_{V_i}$ is a homeomorphism between $V_i$ and $U_z$.
Now each $y_i$ likewise has an evenly covered (for $q$) neighbourhood $W_{y_i}$.
We can now replace $U_z$ by the better neighbourhood $U'_z:= \cap_{i=1}^n r[W_{y_i} \cap V_i]$ (which is open as a finite intersection of open sets). $U'_z$ is still evenly covered by $r$ but now the $y_i$ disjoint neighbourhoods are $W_{y_i} \cap V_i$ (and these in turn are still evenly covered by $q$).
You can now easily check that $U'_z$ is the required evenly covered neighbourhood for $p$. See this answer for this same idea with slightly more details, which I found later.
