This is example 8 from Spanier section 2.4. The example is as follows, we have $X$ to be the union of $$A_1 = \{(x, y)|x=0, -2\leq y\leq 1\}\\ A_2 = \{(x, y)|0\leq x\leq 1, y=-2\}\\ A_3 = \{(x, y)|x=1, -2\leq y\leq 0\}\\ A_4 = \{(x, y)|0<x\leq 1, y=\sin 2\pi/x\}$$ as a subspace of $\mathbb{R}^2$. Next, he takes $\tilde{X}$ to be $[0, 4)$ and maps $[0, 1], [1, 2], [2, 3]$ linearly onto $A_1, A_2, A_3$ respectively and maps $[3, 4)$ homeomorphically (he says $t\mapsto (t-3, \sin (2\pi/(t-3)))$ but it's supposed to be $4-t$?) onto $A_4$. He then claims that this is a fibration with unique path lifting.
I can see why there is unique path lifting, and how to lift maps from $Y$ if $Y$ is locally path connected, but how do I show that the map is a fibration in general? Because $\tilde{X}$ is contractible, any $f\colon Y\to X$ that lifts to $\tilde{X}$ must be nullhomotopic. So, the question is how to lift a null homotopy from $X$ to $\tilde{X}$? What do null homotopic maps to $X$ look like?
A proof does not seem to be easy. I found a general theorem saying that it is true in
See here Theorem 6.5.
Note that the space $X$ is a variant of the Warsaw circle.
Perhaps one can simplify the proof in the example of your question, but I doubt that we get a really short proof.