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.

Consider the injective path $g:[0,\infty)\to Y$ starting at $(0,0)$, running through the arc connecting $(0,0)$ and $(1/\pi,0)$, and sending $x$ to $(1/x,\sin x)$ for $x\ge\pi$. Assume that $\tilde f:Y\to R$ is a lift for $f:Y\twoheadrightarrow S^1$. As $f$ is injective except for the vertical segment, and $g$ avoids this segment, $\tilde fg$ must be injective, and thus strictly increasing. Denote the point $\tilde fg(0)=\tilde f(0,0)$ by $y_0$. Consider the points $x_n=\left(\frac1{n\pi},0\right)$. We have $x_n=g(n\pi)$, so the sequence $y_n=\tilde f(x_n)$ increases. But $(x_n)_n\to (0,0)$, so $y_n\to y_0$, which is impossible as $y_0=\tilde fg(0)$ is less than $y_1=\tilde fg(\pi)$.