Suppose I have a covering map $p:X\rightarrow Y$, and $Y$ is connected. Is $Y$, as an open set, evenly covered by $p$?
I think the answer is yes; I'm new to this kind of topology, so I'm not sure if this proof works: If $Y$ is connected, then there will be at least two points $y_0,y_1$ with intersecting connected, open neighbourhoods that are evenly covered. Suppose these neighbourhoods are $U_0,U_1$ and suppose $\{V_\alpha\}$ is the partitioning of $p^{-1}(U_0)$ and $\{W_\beta\}$ is the partitioning of $p^{-1}(U_1)$. Then since $U_0\cap U_1$ is open, and every partition is homeomorphic, then $p^{-1}(U_0\cap U_1)\cap V_\alpha\neq \emptyset$, and $p^{-1}(U_0\cap U_1)\cap W_\beta\neq\emptyset$, for all $\alpha$ and $\beta$. Then for each $V_\alpha$, there is a single $W_\beta$ such that $p(V_\alpha\cap W_\beta)=U_0\cap U_1$. Since the union $V_\alpha\cup W_\beta$ is an open set that is homeomorphic to $U_0\cup U_1$, and there is a unique way to match the two partitions up, the new partition $\{V_\alpha\cup W_\alpha\}$ creates an even covering of $U_0\cup U_1$; repeating this process over the neighbourhoods of all points will eventually create an even covering of the the open set $Y$.
Does that work?
This doesn't work; the problem is that a single $V_\alpha$ might intersect more than one $W_\beta$. In particular, if $U_0\cap U_1$ is disconnected, there might be one $\beta$ such that $V_\alpha\cap W_\beta$ maps homeomorphically to one component of $U_0\cap U_1$, and a different $\beta'$ such that $V_\alpha\cap W_{\beta'}$ maps homeomorphically to another component of $U_0\cap U_1$. So there is no single $V_\alpha\cup W_\beta$ that maps homeomorphically to all of $U_0\cup U_1$.
To see how this can happen concretely, try to figure out what happens when you apply your argument the case where $X=Y=S^1$ and $p$ is given by $p(z)=z^2$, thinking of $S^1$ as the unit circle in $\mathbb{C}$. You can take $U_0$ to be $S^1\setminus\{1\}$ and $U_1$ to be $S^1\setminus\{-1\}$.