The setting is that manifolds are Banach manifolds, not necessarily finite dimensional. No other assumption is made about the topology of the manifold. In particular, it is not assumed to be regular or normal. Of course, this means that the manifold is not assumed to be paracompact. There is no assumption made about the manifold admitting partitions of unity. In this setting, both Lang (Fundamentals of Differential Geometry, 1999, Springer-Verlag) and Abraham, Marsden, and Ratiu (Manifolds, Tensor Analysis, and Applications, 1988, Springer-Verlag) prove that a connected Hausdorff manifold with a Riemannian metric is a metric space. Both proofs, it seems to me, suffer from the same flaw. The proofs are similar enough that I'll just refer to Lang's. We start with a connected Hausdorff manifold $X$ and a Riemannian metric, $g$ on $X$. No special assumptions are made about the Hilbert space, $E$, on which the manifold is modeled. For example, $E$ may or may not be separable. Start by defining a length function, $L_g$ which assigns a real number $L_g(\gamma)$ to each piecewise $C^1$ path, $\gamma$, from $J=[a,b]$ into $X$, based on the metric, $g$. The distance function, $d_g:X\times X\to\mathbb{R}$ is then defined by $d_g(x,y)=\inf\{L_g(\gamma)\}$ over all piecewise $C^1$ paths, $\gamma$, defined on $J$ with $\gamma(a)=x$ and $\gamma(b)=y$.

Without any difficulty, $d_g$ is a pseudo-metric. The first main point of the proof is to show that $d_g$ is actually a metric. So we start with distinct points $x$ and $y$ of $X$ and set out to show $d_g(x,y) > 0$. We have a chart $(U, \phi)$ at $x$ for $X$ with $\phi(U)$ open in $E$, and we can arrange $U$ to be small enough that $y$ is not in $U$, since the manifold is assumed to be Hausdorff. Working in $\phi(U)$ we find an $r>0$ such that the closed ball $D(\phi(x),r)$ is contained in $\phi(U)$ and such that certain other properties hold. Let $S(\phi(x),r)$ be the boundary of $D(\phi(x),r)$. Then we define $D(x,r)=\phi^{-1}(D(\phi(x),r))$ and $S(x,r)=\phi^{-1}(S(x,r))$, both subsets of $U$.

Since $\phi$ is a homeomorphism, $D(x,r)$ and $S(x,r)$ are closed in $U$ (not necessarily closed in $X$). To me, this is a key stumbling point, as I'll explain. We next let $\gamma:J \to X$ be any piecewise $C^1$ path in $X$ from $x$ to $y$. Both proofs make the following assumption: since $x$ is in $D(x,r)$ and since $y$ is not in $U$, the path $\gamma$ must cross $S(x,r)$. Neither author explicitly proves this assumption (and AMR doesn't even state it).

When I set out to prove this, using the continuity of $\gamma$ and the connectedness of $J$, I quickly run into the need to show that $D(x,r)$ is closed in $X$, not just in $U$. If $X$ were known to be regular, it would not be a problem to take $r$ small enough that $D(x,r)$ was closed in $X$. But as I mentioned at the beginning, I don't know that $X$ is regular. If I could show that the pseudo-metric topology for $X$ induced by $d_g$ was the same as the original manifold topology, I would also get that $X$ was regular. But I don't see how to do that without first completing the first part of the proof.

The whole question seems to be, can I make $r$ small enough that $D(x,r)$ stays away from the topological (in the original manifold topology of $X$) boundary of $U$? But this does not seem to be a local issue, since it depends on what is closed in $X$ which in turn, depends on what is open everywhere in $X$ including outside of $U$.

So, the question is, are other assumptions necessary, or is it possible to fix the proof so that no other assumptions need be made?