Covering map implies local homeomorphism

Statement

Suppose $E$ and $B$ are topological spaces and $p:E\to B$ is a Covering map (?). Then, $p$ is a Local homeomorphism (?).

Proof

Given: A covering map $p:E\to B$. A point $e\in E$.

To prove: There exists an open subset $V$ of $E$ containing $e$ such that the restriction of $p$ to $V$ is a homeomorphism.

Proof:

Step no.	Assertion	Definitions used	Previous steps used	Explanation
1	Let $b=p(e)$	--	--
2	There exists an open subset $U$ of $B$ such that $b\in U$, a discrete space $F$, and a homeomorphism $\varphi :U\times F\to p^{-1}(U)$ such that $p\circ \varphi$ is projection on the first coordinate.	covering map	(1)
3	Let $f$ be the second coordinate of $\varphi ^{-1}(e)$. Let $V$ be $\varphi (U\times \{f\})$. Then $V$ contains $e$	--	(2)	By definition, $\varphi ^{-1}(e)\in U\times \{f\}$. Since $\varphi$ is bijective, we get that $e\in \varphi (U\times \{f\})$.
4	$V$ is an open subset of $p^{-1}(U)$	--	(2), (3)	Since $\varphi$ is a homeomorphism, it suffices to note that $U\times \{f\}$ is an open subset of $U\times F$. This in turn follows from the fact that $F$ being a discrete space, $\{f\}$ is an open subset of $F$.
5	$V$ is an open subset of $E$	--	(2), (4)	Since $p$ is continuous and $U$ is open in $B$, $p^{-1}(U)$ is open in $E$. Step (4) says that $V$ is open in $p^{-1}(U)$. Combining, we get that $V$ is open in $E$.
6	The restriction of $p$ to $V$ gives a homeomorphism from $V$ to $U$.	--	(2)-(5)	We have that $p=(p\circ \varphi )\circ \varphi ^{-1}$. Restricted to $V$, the map is the composite of a homeomorphism from $V$ to $U\times \{f\}$ and a projection (homeomorphism) from $U\times \{f\}$ to $U$. Hence, the map overall is a homeomorphism.