The Problem: Let $p: E\to B$ be a covering map. If $B$ is completely regular, so must $E$.
The question has been asked before here and here. The main thing I am confused about is, for example, the answer here says
In fact, $f'=f\circ p$ not only on $V_\beta\cap U$ but also outside of $p^{-1}(V)$ and on $\mathbf{V_\beta - U}$, because both functions equal $0$. So, they are equal and continuous on the closed set $E-\cup_{\alpha\neq\beta}V_\alpha$. Besides, $f'\equiv 0$ and is continuous on the closed set $E-V_\beta$. The two closed sets cover the whole space $E$, so that, $f'$ is continuous.
My Question: Why does $f\circ p=0$ on $V_\beta - U$? I understand that $p$ is a homeomorphism on $V_\beta$ and thus is bijective; but how do we know then that there does not exist some $x\in V_\beta-U$ such that $p(x)\in p(U)$? Any help would be greatly appreciated.
Your doubts are justified, it is possible that $p(x) \in p(U)$. As an example take $B = \mathbb R$, $E = B \times \{0, 1\}$ and $p : E \to B$ the projection onto $B$. Consider the point $x = (2,0) \in E$. Then $y = p(x) = 2$. Let $U = (1,3) \times \{0\} \cup B \times \{1\}$ and $V = (0,3)$. Then $p(U) = B$ and $V \cap p(U) = V$. For $f$ we can take a function which is $0$ on $B \setminus V$, positive on $V$ and $1$ at $y = 2$.
Clearly $V$ is evenly covered with two sheets $V \times \{0\}$ and $V \times \{1\}$. We have $x \in V \times \{0\}$. But $V \times \{0\} \setminus U$ is non-empty and $p(V \times \{0\} \setminus U) \subset V \subset B = p(U)$. Now recall that $f$ is positive on $V$.
This problem is caused by an unfortunate argument in the linked proof. Here is what we can do better.
We want to find a function $f' : E \to [0,1]$ such that $f'(x) = 1$ and $f'(x') = 0$ for $x' \notin U$.
Consider an evenly covered open neighborhood $V$ of $y = p(x)$ in $B$. Now let $U' = V_\beta \cap U$. Then $p(U') \subset V$, thus also $V' = p(U')$ is an evenly covered open neighborhood of $y$ in $B$ with sheets $V'_\alpha = V_\alpha \cap p^{-1}(V')$. Clearly $V'_\beta = U'$.
Since $B$ is regular, there exist an open neigborhood $W$ of $y$ whose closure is contained in $V'$. Now choose $f: B \to [0,1]$ such that $f(y) = 1$ and $f(y') = 0$ for $y' \notin W$. Define $$f' : E \to [0,1], f(x') = \begin{cases} f(p(x')) & x' \in U' \\ 0 & x' \in S = (E \setminus p^{-1}(\overline W)) \cup \bigcup_{\alpha \ne \beta} V'_\alpha \end{cases}$$ Let us show that this is well-defined.
$U' \cup S = E$ because $U' \cup S = V'_\beta \cup S = (E \setminus p^{-1}(\overline W)) \cup \bigcup_{\alpha} V'_\alpha = (E \setminus p^{-1}(\overline W)) \cup p^{-1}(V') = E$.
$U' \cap S = V'_\beta \cap S = V'_\beta \setminus p^{-1}(\overline W)$.
For $x' \in U' \cap S$ we have $f'(p(x')) = 0$: In fact, if $x' \in U' \cap S = V'_\beta \setminus p^{-1}(\overline W)$, then $x' \notin p^{-1}(\overline W))$, thus $p(x') \notin \overline W$ and in particular $p(x') \notin W$. Thus $f'(p(x')) = 0$.
Since $S$ is open, we conclude that $f'$ is continuous. By definition $f'(x) =1$ and $f'(x') = 0$ for $x' \notin U'$ (note that $E \setminus U' \subset S$ by 1.). But $U' \subset U$, thus $f'(x') = 0$ for $x' \notin U$.