Given an $n$-dimensional manifold $M$ with Riemannian metric $\eta$, the orthonormal frame bundle over $M$ is defined as the set of all orthonormal frames over all points in $M$. Denote such bundle by $O(M)$.
Claim: It is a trivial bundle over $M$
"Proof": For every $x \in M$, fix an orthonormal basis for $T_xM$. Let $\phi: O(M) \to O(n, \mathbb{R})\times M$ be defined by $\phi(r_x) = (A_{r_x}, x)$, where $A_{r_x}$ is the matrix whose columns are $X_1, ..., X_n$, where $r_x = (X_1, ..., X_n)$ and the coordinates are written with respect to the fixed basis.
Now, if we define $\theta: O(n, \mathbb{R})\times M \to O(M)$ by $\theta(A, x) = (r_x)$, where $r_x = (A_1, ..., A_n)$, the columns of $A$, in the tangent space $T_xM$, we get that $\theta$ and $\phi$ are inverses of each other.
Thus, $O(M)$ inherits a manifold structure via the bijection $\phi$ and, therefore, this mapping becomes a diffeomorphism. $\square$
Question: What is wrong with the above proof? I'm not very good with geometry, and I can't see the flaw in the reasoning...
Thanks in advance!
EDIT: Given @Didier's comment, my suspicions were indeed correct and the above isn't a trivialization. In that case, what is the topology on $O(M)$ that makes it a manifold?