The question really is that simple:

Prove that the manifold $SO(n) \subset GL(n, \mathbb{R})$ is connected.

it is very easy to see that the elements of $SO(n)$ are in one-to-one correspondence with the set of orthonormal basis of $\mathbb{R}^n$ (the set of rows of the matrix of an element of $SO(n)$ is such a basis).

My idea was to show that given any orthonormal basis $(a_i)_1^n$ in $\mathbb{R}^n$ there's a continuous deformation from $(a_i)_1^n$ to $(e_i)_1^n$ the usual basis passing only through orthonormal basis. Such a deformation would yield a path between any element of $SO(n)$ and $I$, and the theorem would follow.

Also, a geometric picture is also "simple", one would simply spin the first basis until $a_1$ agrees with $e_1$ and proceed from there. However i'm having a lot of trouble coming up with this spinning process.

Any help would be appreciated.