Is $f$ necessarily a covering? Let $f : X \rightarrow Y$ be a continuous map of spaces where $X$ is compact Hausdorff, $Y$ is Hausdorff and both spaces are path-connected and locally path-connected. Suppose
that for every $x \in X$ there exists an open neighborhood $U$ such that $f(U)$ is open in
$Y$ and $f|U : U \rightarrow f(U)$ is a homeomorphism. Is $f$ necessarily a covering?

Some remarks that might help: 
First, $f$ is clearly an open map, because it is a local homeomorphism. So $f[X]$ is open, compact (and so closed, as $Y$ is Hausdorff) in a (path-)connected space, so $f[X] = Y$ and $f$ is surjective.
Also, for every $x \in X$, let's call the $U$ from the assumptions $U_x$, so that $f$ restricted to $U_x$ is a homeomorphism between open sets. If $y$ is in $Y$, then $y = f(x)$ for some $x$, and for every such $x$, $U_x$ contains no other points of $F_y:= f^{-1}[\{y\}]$, the fibre of $y$, because $f$ is injective on $U_x$. So for every $x \in F_y$, $U_x \cap F_y = \{x\}$, so $F_y$ is a discrete subspace of $X$, and as it is a closed discrete subspace in a compact $X$, it is finite. So all fibres are finite.
The final part is provided in this answer. 
Noteworthy: no local connectedness is needed in either space. Connectedness only to get surjectivity.  
