prove that $M_xM_1 = ?$ Let $\mathbf{y} = \mathbf{X_1}\mathbf{\beta_1}+\mathbf{X_2}\mathbf{\beta_2}+\mathbf{u}$. 

Let $\mathbf{X} = [\mathbf{X_1} |\mathbf{X_2}]$.
Define $\mathbf{M_i}$ to be the complementary projections for $i = 1,2$. 

Then, $\mathbf{M_xM_1} = \mathbf{M_x}(\mathbf{I} - \mathbf{P_1}) = \mathbf{M_x - M_xP_1} = \mathbf{M_x-[I-P_x]P_1} = \mathbf{M_x-[P_1-P_xP_1]} = \mathbf{M_x - [P_1-P_1]} = \mathbf{M_x}$. 

Could you check if this is okay?
The only assertion here that goes beyond self-evident matrix algebra and substitutions, is that 

$$P_XP_1 = P_1$$

Since the columns of $X_1$ are in $X$ it follows that

$$P_X X_1 = X_1$$

because in general it holds that 

$$P_X X = X$$

Namely the projection matrix operates as the identity matrix for the columns of the matrix from which it has been constructed. But then

$$P_X P_1 = P_XX_1(X_1'X_1)^{-1}X_1' = (P_XX_1)(X_1'X_1)^{-1}X_1'$$

$$=X_1(X_1'X_1)^{-1}X_1' = P_1$$