Function doesn't have a lift in a space related to Topologist's sine curve

I'll use the convention that $S^1 = \{z\in\mathbb C : \lvert z \rvert = 1\}$ and $p(t)=e^{2\pi i t}$, to avoid the clash of notations between ordered pairs in $\mathbb R^2$ and open intervals in $\mathbb R$.

Let $L$ be the segment on the $y$-axis. WLOG assume that $f(L)=\{1\}$. Suppose $\tilde f\colon Y\rightarrow R$ is a lift. Since $Y\setminus L$ is connected, $\tilde f(Y\setminus L)$ is also connected, so it must lie in one of the components of $p^{-1}(f(Y\setminus L))=\mathbb R \setminus 2\pi \mathbb Z$, say the interval $(0,2\pi)$. It follows by the surjectivity of $f$ that $\tilde f(Y\setminus L)$ must be precisely $(0,2\pi)$. Now $Y$ is compact, so $\tilde f(Y) \supset [0,2\pi]$. Thus, $\{0,2\pi\}\subset \tilde f(L)$, contradicting your observation that $\tilde f(L)$ is one point.

Alternatively for the last step, observe that $\tilde f(L)$ should be connected, but $\tilde f(L) \subset 2\pi\mathbb Z$ is a discrete set containing at least two points.