Let $p: E \to B$ be a covering map. If $B$ is a completely regular space then prove that $E$ is completely regular space. I am getting no clue how to construct the function $f$.
The readers may check the same problem here Is a covering space of a completely regular space also completely regular
No answer is there also.
You don't need to use exactly a pullback $f$ (in fact you can't). You actually have a good deal of flexibility in constructing functions. Try to visualize the function you'd want, then prove that you can make a function like that.
Here's a hint to get you going: Given $x \in E$ and $V \subset E$ closed not meeting $x$, find an open neighborhood $N$ of $x$ small enough to both miss $V$ and satisfy the covering condition, by which I mean p(N) satisfies the covering condition (why can you do this?).
Then construct a function on $B$ which is $0$ at $p(x)$ and is $1$ on $B - p(N)$.
You'll want some nested neighborhoods in what follows, so now take $N$ smaller, say by intersecting with the set where p composed with this function is less than 1/2, and then construct this type of function again.
Now modify the pullback of the function near other lifts of p(N) to get you what you want and prove your modification is still continuous.