Let $f:X\rightarrow S^1$ be a continuous function for $X$ connected and $x^* \in X$. Given that $\hat{f}_1, \hat{f}_2:X \rightarrow \mathbb{R}$ are liftings of $f$ such that $\hat{f}_1(x^*) = \hat{f}_2(x^*)$, I am trying to prove that $\hat{f}_1(x) = \hat{f}_2(x)$ for all $x \in X$.
$\hat{f}_1 - \hat{f}_2$ is a continuous function, and $(\hat{f}_1 - \hat{f}_2)(X)$ is a connected set in $\mathbb{R}$. I'm trying to prove the aforementioned result, but I'm having a bit of difficulty. It would suffice to show that $(\hat{f}_1 - \hat{f}_2)(x) = 0$ for all $x$, but I don't necessarily know how to show the image of $X$ under this function can't be a connected interval in $\mathbb{R}$. Any suggestions?
Hint: consider the set of points in $X$ where the two lifts coincide. It is nonempty and obviously closed. To show it is open, use the definition of covering space.