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)
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 $M$.