Yes, indeed! It fails exactly at $x_0$ (and nowhere else).
To show this, look at the inverse image of a small open neighbourhood $U$ of $x_0$. Intuitively, it should be clear that this can’t be expressed as a disjoint union of copies of $U$. To show this rigorously, first do a warm-up exercise (or recall it, if you’ve seen it before): show that the letter X is not homeomorphic to an interval.
That approach shows very concretely how the definition of covering space fails here. If you have already seen the path lifting property for covering spaces, then that can be used to give a nice, less direct proof that this example is not a covering space. Just look at a path going through $x_0$ the “wrong” way…