covering projection If f:E->B is a covering projection & B is completely regular then how E is also completely regular?
 A: This is Exercise 6 on page 341 of Munkres's Topology (Second Edition). Indeed, if $B$ is Hausdorff, regular, completely regular, or locally compact Hausdorff then so is $E$. Here is the hint Munkres gives:
If {$V_\alpha$} is a partition of $p^{-1}(U)$ into slices and $C$ is a closed set of $B$ s.t. $C\subset U$, then $p^{-1}(C)\cap V_\alpha$ is a closed set of $E$.
Let's follow the hint. To prove $E$ is completely regular when $B$ is completely regular, it's enough to show that for any $e \in E$ and any neighborhood $V$ there is a continuous function $g : E \rightarrow [0,1]$ s.t. $g(e) = 1$ and $g(x)=0$ for all $x\not \in V$. If this holds then certainly for any closed set $C$ not containing $e$ we'll have a function taking $C$ to zero and $e$ to $1$, simply setting $V = E-C$ above.
To construct $g$, let $b = p(e)$ and $U = p(V)$. We may assume $U$ is evenly covered and $V$ is a slice above $U$, because if $U$ is not then simply take a subneighborhood of $b$ which is evenly covered. Because $B$ is completely regular, there is a map $f: B \rightarrow [0,1]$ s.t. $f(b) = 1$ and $f(x)=0$ for all $x\not \in U$. Let $W$ be the union of the slices other than $V$. 
Define $g$ to equal $f \circ p$ on $V$ and to be $0$ on $E−V$. Then $g$ is equal to $f \circ p$ on $E−W$ because $f \circ p$ is $0$ outside $p^{−1}(U)$. Hence $g$ is continuous on the closed sets $E−V$ and $E−W$, and therefore on all of $E$.
