Do Riemannian manifolds have to be smooth to define a metric or can it be relaxed to differentiable, or even simply just a path-connected set? If I have a path-connected (closed) differentiable manifold $M$ naturally in $\mathbb{R}^n$, for example a path-connected component of the level set $h = 0$ with $h \colon \mathbb{R}^n \to \mathbb{R}^m$ continuously differentiable such that $\mathrm{d}h$ is surjective on all $\mathbb{R}^n$, can I endow it with a metric? I am not looking for smooth properties, simply the notion of a distance.
I am guessing I can define the length of a continuous path as its length in $\mathbb{R}^n$ and the distance between two points $x,y \in M$ as the infimum of the length of continuous paths on $M$ from $x$ to $y$. Would that make sense? (although differentiability here is not used)
 A: It all depends on your definition of "manifold" (or "closed manifold"). If we use the standard terminology, it is correct to say that every paracompact differentiable manifold has a Riemannian metric (and Riemannian metric then defines a distance function on every connected component). See Wikipedia.
I think it does not make much sense to talk about the length of a continuous path if the path in question is not continuously differentiable. The reason is that there exist continuous paths that do not have finite length in the standard sense of the word (so some care should be applied when you define the infimum).
A less precise point: you ask if Riemannian manifolds have to be smooth to define a metric. There are so-called symplectic structures, to define which you need the manifold to be smooth. However, people also consider limits of diffeomorphisms preserving the symplectic structure that are not themselves differentiable (such continuous maps are named "symplectic homeomorphisms", see here) and their dynamics is apparently pretty interesting. Possibly, one can do something similar with Riemannian structures but I am not sure (Riemannian structure is more rigid than the symplectic one so I doubt it but who knows).
A: There is a general definition of the length of any continuous curve $\gamma : [0,1] \to \mathbb R^n$, although it might be infinite. Namely, $\text{Length}(\gamma)$ is the supremum, over all partitions
$$0 = x_0 < x_1 < ... < x_{K-1} < x_K = 1
$$
of the quantity
$$|\gamma(x_0)-\gamma(x_1)| \, + \, ... \, + \, |\gamma_{K-1} - \gamma_K|
$$
It's not too hard to prove that if $\gamma$ is a piecewise $C^1$ curve then $\text{Length}(\gamma) < \infty$; you can find that in many advanced calculus books.
If $M \subset \mathbb R^n$ is a connected $C^1$-differential manifold, it follows that it is path connected, and furthermore that any two points $x,y \in M$ are the endpoints of some piecewise $C^1$ path (one can actually get a $C^1$ path with a tiny bit more work, but it doesn't simplify anything else to do that). 
One can now define $d(x,y)$, for any $x,y \in M$, to be the infimum of $\text{Length}(\gamma)$ taken over all $C^1$ paths $\gamma$ having endpoints $x,y$. And it's not too hard to prove that this is indeed a metric on $M$.  
