Mathematical proof of a combined hash function
Ignoring the issue of output length I mention in the comments, you should be able to prove this without intermediate steps. Here is an outline of how I would do it.


Without loss of generality let one particular hash function be the one that is collision resistant. It doesn't matter which one. Say $H_3$.
Assume that $M$ is not collision resistant. This means, you can find $m_0\neq m_1$ such that $M(m_0)=M(m_1)$.
Show how this would lead to a collision in $H_3$ directly. This leads to a contradiction, so $M$ must be collision resistant. 


This is a proof by contradiction.