So I just got through proving the following theorem:
If $p:C\to X$ is a covering map and $Y$ is a [xxx] space, then given $y_0\in Y$, $c_0\in C$, $f:Y\to X$ such that $f(y_0)=p(c_0)$ there exists at most one lift $\hat f:Y\to C$ such that $p\circ\hat f=f$ and $\hat f(y_0)=c_0$.
Adding assumptions as they arose in the proof, I naturally ended up assuming that the [xxx] says $Y$ is connected and weakly locally connected (not path-connected). Now obviously $Y$ has to be connected, since otherwise you can "jump" from one point in the fiber to the next along a disconnection, for example if $C=\Bbb R$, $X=\Bbb R/\Bbb Z$ and $Y=\{0,1\}$, if $y_0=c_0=0$ and $f(0)=f(1)=0$ then $\hat f(0)=0$, $\hat f(1)=n$ is a lift for any $n\in\Bbb Z$.
But I'm having much more difficulty convincing myself that we have to assume that $Y$ is also weakly locally connected (recall this means that every point has a local base of connected but not necessarily open sets). Question: Is the assumption that $Y$ is weakly locally connected necessary?
Proof: Suppose $\hat f,\hat f'$ are lifts. Then for each point $y\in Y$ there is a neighborhood $U\subseteq X$ of $p\circ \hat f(y)$ which is evenly covered, so there is some homeomorphic preimage $\hat f(y)\in V\overset p{\simeq} U$, and then by local connectivity there is a connected $y\in S\subseteq \hat f^{-1}(V)$, then if $z_1,z_2\in S$ such that $\hat f(z_1)=\hat f'(z_1)$, then since $S$ is connected and $\hat f'(z_1)=\hat f(z_1)\in V$, $\hat f'$ maps all of $S$ to $V$ (so $\hat f'(z_2)\in V$), and since $z_2\in S\subseteq \hat f^{-1}(V)$ as well $\hat f(z_2)\in V$. Thus since $p$ is locally an injection on $V$, $p\circ\hat f(z_2)=p\circ\hat f'(z_2)$ implies $\hat f(z_2)=\hat f'(z_2)$.
We have shown that there is a neighborhood $S$ of each point $y\in Y$ with $\hat f(z_1)=\hat f'(z_1)$ iff $\hat f(z_2)=\hat f'(z_2)$ for all $z_1,z_2\in S$, so the set of all $y\in Y$ such that $\hat f(y)=\hat f'(y)$ is clopen, and connectivity of $Y$ and $\hat f(y_0)=\hat f'(y_0)$ gives $\hat f=\hat f'$.