I was reading the proof of the --well known-- theorem that if $(M, g)$ is a Riemannian manifold, then, one defines the Riemannian distance $d$ on $M$, as $$ M \times M \ni (a, b) \longmapsto d(a, b) = \inf \left\{ L(c) \bigm| c \textrm{ is a $C^1$ piecewise curve that joins } a, b \right\}$$ where $L(c) = \int_I \sqrt{g(c(t))\cdot(c'(t), c'(t))} \; \textrm d t$. The theorem states that $d$ is indeed a metric compatible with the natural topology of $M$. My question is about the proof found in [1, VII.6] and [2, V.3.1] at the point where it asserts that " $a \neq b$ implies $d(a,b) > 0$. "
It mentions that (*) $a \neq b$ implies that we can find a local chart $(U, \varphi)$ and a closed ball $B$ in $\varphi(U)$ centered at $\varphi(a)$, of radius $r > 0$ such that any curve joining $a$ and $b$ should intersect the sphere $\varphi^{-1}(\partial B)$.
I cannot find any proof to this fact, unless i put the constraint of regularity in the manifold (so it turns to a $T_3$ space). Ok, i can understand that in the case that $M$ is locally compact manifold, or even paracompact, then Hausdorff property is sufficient, but even then, this fact, needs a proof based on the local topological properties of $M$.
My question is: Can we prove (*) without further constraints?
My references to this question are:
[1] S. Lang, "Introduction to Differential manifolds"
[2] W. Boothby, "Introduction to Differential manifolds and Riemannian Geometry"