How to show the residual sum of squares of one model is at least the residual sum of squares of another? 
This question has really been bugging me and I'm failing to understand it conceptually. I don't see why it is the case and would really appreciate any help.
 A: This is important:

all predictors in the first model are also contained [in] the second model.

So let's just assume that $X_1$ contains the first columns of $X_2$ (we can always reorder the columns of $X_2$ otherwise, this won't change anything.
Consider the solution $\gamma_1$ of the first model, which yields some $SS_{\text{Res},1}$.
If we fill $\gamma_1$ up with zeros to match the dimensionality of the second model, we get a feasible solution to the second model. Note that the $SS$ of this solution is exactly $SS_{\text{Res},1}$, since all additional predictors enter with zero coefficients.
However, we now have the freedom to change the coefficients of the second model to yield an even better fit $\beta$, i.e., one with $SS_{\text{Res},2}<SS_{\text{Res},1}$. This may or may not be possible, but we will always have at least $SS_{\text{Res},1}\geq SS_{\text{Res},2}$.
Conceptually: since the larger model encompasses the smaller one, the optimum fit for the smaller model will always be feasible for the larger one, so $SS$ will always go down (or stay constant) if we enlarge the model.
