There are two differences between what this question is asking and standard results that are easy to find in the literature: (a) We don't assume ahead of time that $W$ is finite, but only that it has finitely many reflections, and (b) We don't assume ahead of time that the reflections all fix some Euclidean metric. We can get these assumptions back without much trouble:

(a) Here is a direct argument that "finitely many reflections" implies "finite". *See the edit history of this question for a more complicated argument with a gap. (Thanks to the OP for pointing out the gap.) The nature of the gap leads to the key point in this new argument.*

Assume that $W$ contains finitely many reflections and consider the hyperplane arrangement consisting of its reflecting hyperplanes. The *regions* of the arrangement are the closures of connected components of the complement.

The action of $W$ takes regions to regions, because the facets of regions are defined by hyperplanes and the action of $W$ takes reflecting hyperplanes to reflecting hyperplanes. (Given a hyperplane $H_t$ and an element $w\in W$, the hyperplane $wH_t$ is $H_{wtw^{-1}}$.) If $W$ is infinite, then each region $R$ has an infinite stabilizer $W^R$, and since $R$ has finitely many extreme rays and elements of $W^R$ permute the extreme rays, there are infinitely many elements of $W$ that fix each extreme ray of $R$ setwise.

In particular, let $w$ be a nontrivial element of $W$ that fixes each extreme ray of $R$ pointwise. If $R'$ is an adjacent region to $R$, then since $w$ fixes their common facet and sends regions to regions, $w$ fixes $R'$ as a set as well, and thus fixes every ray of $R'$ as a set. We conclude that $w$ fixes every ray of every region.

Said another way, every ray consists of eigenvectors of $w$ (with positive, real eigenvalues). Furthermore, taking nonzero vectors from different rays of $R$, we obtain a basis of eigenvectors. Thus the ambient space decomposes as a direct sum of eigenspaces. If there is only one eigenspace, then $w$ scales the whole space by the eigenvalue, which is not $1$, since $w$ is not the identity. This is a contradiction, because the determinant if $w$ is not $\pm1$, but $W$ is generated by elements with determinant $-1$.

If there are multiple eigenspaces, then since the rays are contained in the eigenspaces and the reflecting hyperplanes are spanned by rays, each reflecting hyperplane contains all but one of the eigenspaces and intersects the other in codimension $1$. We see in this case that $W$ is a direct product of factors, each of which is isomorphic to a group generated by reflections on one of the eigenspaces. By induction on dimension, each of these is finite.

(b) There is a standard argument that for any finite group of linear transformations, there is a Euclidean symmetric bilinear form preserved by the group.

In case you have trouble finding the exact statements you want:

I have a chapter *Finite Coxeter Groups and the Weak Order* in the book Lattice Theory: Special Topics and Applications, Volume 2, (ed. George Grätzer and Friedrich Wehrung), Springer 2016 that does these things. When I wrote this, I wrote down some results like this that surely were known but that I didn't know references for (and also many results that I did know references for). It's safe to assume that the results in that chapter *about the basic Coxeter group setup* are not new, but at least it's a place to cite things to if you can't find older results. References below are to that chapter.

The standard argument (b) is given as Proposition 10-2.8 (see also Proposition 10-2.7), but surely this argument is in many references. Once you have run that argument, so that you know you have a Euclidean reflection arrangement, the answer to your question (1) is probably in Bourbaki, or see Theorem 10-2.5. The answer to your question (2) is Theorem 10-2.9. The notes on page 557 cite Theorem 10-2.9 to Coxeter in the 1930's, but I don't recall whether that result is easily quotable from Coxeter or whether it's just implicit in the arguments there.