I'm trying to solve exercise 1.3.7 in Hatcher's Algebraic Topology:
Let $Y$ be the quasi-circle that is the union of a portion of the graph $y = \sin(1/x)$, the line segment $[-1,1]$ in the $y$-axis, and an arc connecting the two pieces. Collapsing the segment in the $y$-axis to a point gives a quotient map $f:Y \to S^1$. Show that $f$ does not lift to the covering space $p:\mathbb{R} \to S^1$, even though $\pi_1(Y) = 0$.
I was able to show that $\pi_1(Y)$ is simply-connected by decomposing it into 2 pieces and showing that a path in $Y$ must be contained in a subspace that is homeomorphic to $\mathbb{R}$. What I'm having difficulty with is showing that $f$ doesn't have a lift. I've shown that if it has a lift $\tilde f$, $\tilde{f}([-1,1])$ is one point. I feel that this contradicts with the continuity of $\tilde f$, but I cannot rigorously prove it.
Any help would be appreciated. This is self-study by the way.