The definition of riemannian metric $g$ requires that the path $\gamma$ is differentiable, and thus the manifold is differentiable. To my understanding it is because that, to find the length of a curve $\gamma$, it needs to be differentiable. So this definition could be generalized to a Lipschitz manifold where the paths have weak derivatives?
What about a metric for a continuous manifold, in general?
This is a conceptual question not homework..
First of all, note that the length of a continuous path $c: [0,1]\to X$ is defined in any metric space $(X,d)$, although some paths will have infinite length: $$ L(c)= \sup \sum_{i=1}^n d(c(t_{i-1}), c(t_i)), $$ where the supremum is taken over all finite increasing sequences $$ 0=t_0< t_1< ...< t_n=1. $$
There are several ways to understand your question. Here is one:
Is it true that every connected topological manifold is "path-metrizable", i.e. admits a compatible metric $d$ such that the distance between any two points is the infimum of length of paths connecting these points?
This question has positive answer. Moreover:
Theorem. Let $X$ be a separable metrizable connected locally connected locally compact topological space. Then $X$ admits a complete geodesic metric, i.e. a metric such that every two points $p, q\in X$ are connected by a geodesic, i.e. a path whose length equals the distance between $p$ and $q$.
See
A. Tominaga and T. Tanaka, Convexification of locally connected generalized continua, J. Sci. Hiroshima Univ. Ser A 19 (1955), 301-306.
Their proof builds upon the earlier work by Moise and Bing.
Connected topological manifolds satisfy the assumptions of this theorem, hence, they admit complete geodesic metrics.
