Does any non-Hausdorff manifold admit a metric tensor of signature $(p,q)$?

Hicks in his book, "Notes on differential geometry", works with manifolds by specifically stating that they are not required to be Hausdorff unless otherwise stated. He then goes on to define the usual manifold structures, vector fields, connections, metric tensors, etc. In chapter 6, he defines the usual length of a curve by a Riemannian metric tensor, that is, for a curve $\gamma$ with tangent vector $T$, the length between the point $\gamma(p)$ and $\gamma(q)$ is

$$|\gamma|^q_p = \int_p^q \sqrt{\langle T(\lambda), T(\lambda)\rangle} d\lambda$$

which he then uses for the definition of a distance function

$$d(p,q) = \inf [|\gamma| , \gamma \text{ piecewise continuous between p and q}]$$

which can be shown to be a pseudometric on the manifold, which specializes to a metric if the manifold is Hausdorff. The problem is, a pseudometric space cannot be $T_0$ and not also a metric space, and every manifold is itself a $T_0$ space. And by the Smirnov metrization theorem, a non-Hausdorff space cannot be a metric space.

I'm not sure exactly where the problem comes from here. Is there a problem of existence of one of the structures involved? Is the metric tensor not continuous in those cases (that is one of the assumption of the proof)? More generally, do such manifolds admit global metric sections for arbitrary signatures, in a different way than Hausdorff manifolds, that is?

• Consider the line with two origins and a path which goes from one origin to the right, then stops and goes to the left to the other origin. Argue that you can make this path as short as you want. This implies that the distance between the two origins is zero. – Thomas Rot Apr 21 '17 at 14:48
• Appealing to the Smirnov metrization theorem to explain the necessity of the Hausdorff property of metric spaces is circular reasoning. – nfdc23 Apr 21 '17 at 14:52
• Taking the line with two origin's example further, as Thomas Rot points out the two different origins in that space are necessarily in any psuedo-metric open ball (centered at one of them say). They cannot be separated by open sets so the topology generated by the pseudo metric is not $T_0$. In particular the pseudometric topology is different from the topology as a non-Hausdorff manifold. – Chris Schommer-Pries Apr 21 '17 at 19:02
• Hicks also claims that the pseudometric topology is equal to the manifold topology in the very same chapter, though, which is why I was rather surprised. – Slereah Apr 22 '17 at 4:40