The group of isometries of a manifold is a Lie group, isn't it? Let $M$ be a connected  finite dimensional topological manifold and $g$ be any metric on it that induces the topology of $M$ ($g$ is not a Riemannian metric). How to prove that the group of isometries of $(M,g)$ is a finite dimensional Lie group?
PS. Note that in case $M$ is a Riemannian manifold, this statement is classical.
 A: The assertion (for connected topological manifolds) is equivalent to the Hilbert–Smith conjecture. The latter (in one of its equivalent versions) asserts that every continuous action of every profinite group on every topological metrizable manifold has an open kernel.
Indeed, let $G$ be a profinite group acting continuously on a connected topological metrizable manifold $X$. Let $d$ be a proper distance on $X$ (inducing the topology). Then $d'(x,y)=\int_G d(gx,gy)$, integrating with respect to the Haar measure, defines another proper distance, continuous on $X^2$, and which in addition is $G$-invariant. Hence it defines the topology and is complete. If OP's assertion holds, then the isometry group being a Lie group, the map has an open kernel.
The Hilbert–Smith conjecture is known in dimension $\le 3$ ($\le 2$ is classical, and $3$ is more recent work of John Pardon).
The assertion is also known to hold when the isometry group acts transitively.

Added remarks on the Hilbert–Smith conjecture:
If "green" qualifies topological groups, say that a topological group $G$ has "no small green subgroups" if there is a neighborhood of the identity $V$ in $G$ such that the only green subgroup of $G$ contained in $V$ is $\{1\}$.
For instance, the solution to Hilbert's 5th problem yields that a locally compact group has no small subgroups iff it is isomorphic (as topological group) to some Lie group.
One reformulation of the Hilbert–Smith conjecture is:

For every connected topological (metrizable) manifold $M$, the topological group $\mathrm{Homeo}(M)$ (with the compact-open topology) has no small compact subgroups.

Newman's theorems (see T. Tao's blog) implies that it has no small finite subgroups. The too optimistic hope would be that $\mathrm{Homeo}(M)$ has no small subgroups, but this is false: as soon as $M$ has positive dimension, it is easy to check that $\mathrm{Homeo}(M)$ has small discrete infinite cyclic subgroups.
