why does the reduced row echelon form have the same null space as the original matrix? What is the proof for this and the intuitive explanation for why the reduced row echelon form have the same null space as the original matrix?

The short answer: Because you multiply nonsingular matrices from the left.
The long answer: Say you have a matrix $A$. Each Gaussian elimination step corresponds to some elementary matrix (which are nonsingular). Thus, there exists a nonsingular matrix $M$ (product of elementary matrices) such that $MA$ has reduced row echelon form. 
Now, let $x$ be in the null space of $A$. Then, we have 
$$MAx = M(Ax) = M0 = 0.$$
That is, $x$ is also in the null space of $MA$. On other hand, let $y$ be in the null space of $MA$. Then, we also have 
$$Ay = M^{-1} (M A y) = M^{-1} 0 = 0.$$ 
Thus, the null spaces of $A$ and $MA$ are the same.
