I am trying a new approach to an already-solved problem, but I need help to see if I'm on point. Munkres Chapter 53, question 6 [abridged] asks, given a covering map $p: E \to B$:
Show that "if $B$ is completely regular, then so is $E$."
Now, first and foremost, I have solved this problem, but I did it via explicit construction of a function $f$ that 'forces' $E$ to be completely regular, given that $B$ is. My question is regarding another possible approach to this proof:
By Munkres's Topology 2nd Ed. Theorem 34.3, a space is completely regular if and only if it is homeomorphic to $[0,1]^{\Omega}$ for some index $\Omega$. Because $B$ is completely regular, we know it is homeomorphic to such an arbitrary product of the closed interval, say for the purposes of this argument that it's product index is $J$. Call this homeomorphism $h$, i.e. $h: B \to [0,1]^J$. Let $E = \bigcup_{\alpha \in K} E_{\alpha}$ be the indexed disjoint union of 'slices' homeomorphic to $B$ under the restriction of $p$ to each slice. But since we know that each 'slice' of $E$ is mapped homeomorphically onto $B$, i.e. for each $\alpha \in K, \space p|_{\alpha} : E_{\alpha} \to B$ is a homeomorphism, and since a composition of homeomorphisms is itself a homeomorphism, each 'slice' $E_{\alpha}$ is homeomorphic to the same closed unit interval product, i.e. let $g_{\alpha} := h \circ p|_{\alpha}: E_{\alpha} \to [0,1]^J$ is a homeomorphism. This shows that each 'slice' is completely regular. Yet we cannot extend $g_{\alpha}$ to a homeomorphism $g: E \to [0,1]^J$ because if we 'generalize' the functions $p|_{\alpha}$, we get the function $p$ (our covering map), which is not necessarily a homeomorphism.
Since we cannot extend $g_{\alpha}$ implicitly, all we need to show then is that a disjoint union of completely regular spaces is completely regular. If so, then the hypothesis is fulfilled, as $E = \bigcup_{\alpha \in K} E_{\alpha}$. It seems like this proposition (almost trivially) answers itself, because every subset of $E$ is either entirely contained in a completely regular slice of some $E_{\alpha}$, hence it is completely regular, or it is the arbitrary union of disjoint completely regular spaces, i.e. by the arguments above, we already know $U \cap E_{\beta}$ is completely regular for any $\beta \in K$. But since each such subspace is completely regular, and the union of these subspaces are the entire set in question, if we can 'build' such a $U$ arbitrarily as the composition of completely regular spaces, does complete regularity of $U$ hold?
Embedded questions notwithstanding, am I on the right track? Does this suffice for a proof? I'm worried that the 'disjoint union' does not preserve complete regularity for an arbitrary $E$ (i.e. I'm stupidly overlooking the obvious...)