Proof that $O_3$ and $GL_3$ are homotopy equivalent
Trying to use polar decomposition to solve this problem and actually I have come to the fact that exists continuous function $sqrt(AA^T) = C$ where A belongs to $GL_n$ and $C$ is positively defined matrix but Im struggling with proving that it is homotopic to the identical
Your idea is on the right road... Yes, for $A\in GL_n$, $AA^t$ has a unique positive definite square root $C$... and $A=C\cdot h$ for an orthogonal matrix $h$. And this decomposition is unique, is part of the conclusion, with or without the explicit description of the $C$ part.
That is, $GL_n \approx P \times O_n$, where $P$ is positive-definite matrices. Since it's actually a Cartesian product, and the cone $P$ is contractible, any choice of such contraction gives the desired homotopy.